### Kinematics: Motion in a Straight Line : Notes and Study Materials -pdf

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### Class 11 Physics Chapter 3 Motion in a Straight Line

Topics and Subtopics in**Class 11 Physics Chapter 3 Motion in a Straight Line****:**

Section Name | Topic Name |

3 | Motion in a Straight Line |

3.1 | Introduction |

3.2 | Position, path length and displacement |

3.3 | Average velocity and average speed |

3.4 | Instantaneous velocity and speed |

3.5 | Acceleration |

3.6 | Kinematic equations for uniformly accelerated motion |

3.7 | Relative velocity |

### Motion in a Straight Line Class 11 Notes Physics Chapter 3

**• Introduction**

Motion is one of the significant topics in physics. Everything in the universe moves. It might only be a small amount of movement and very-very slow, but movement does happen. Even if you appear to be standing still, the Earth is moving around the sun, and the sun is moving around our galaxy.

“An object is said to be in motion if its position changes with time”.

The concept of motion is a re’ live one and a body that may be in motion relative to one reference system, may be at rest relative to another.

There are two branches in physics that examine the motion of an object.

(i) Kinematics: It describes the motion of objects, without looking at the cause of the motion.

(ii) Dynamics: It relates the motion of objects to the forces which cause them.**• Point Object**

If the length covered by the objects are very large in comparison to the size of the objects, the objects are considered point objects.**• Reference Systems**

The motion of a particle is always described with respect to a reference system. A reference system is made by taking an arbitrary point as origin and imagining a co-ordinate system to be attached to it. This co-ordinate system chosen for a given problem constitutes the reference system for it. We generally choose a co-ordinate system attached to the earth as the reference system for most of the problems.**• Total Path Length (Distance)**

For a particle in motion the total length of the actual path traversed between initial and final positions of the particle is known as the ‘total path length’ or distance covered by it.**• Types of Motion**

In order to completely describe the motion of an object, we need to specify its position. For this, we need to know the position co-ordinates. In some cases, three position co-ordinates are required, while in some cases two or one position co-ordinate is required.

Based on these, motion can be classified as:

(i) One dimensional motion. A particle moving along a straight-line or a path is said to undergo one dimensional motion. For example, motion of a train along a straight line, freely falling body under gravity etc.

(ii) Two dimensional motion. A particle moving in a plane is said to undergo two dimensional motion. For example, motion of a shell fired by a gun, carrom board coins etc.

(iii) Three dimensional motion. A particle moving in space is said to undergo three dimensional motion. For example, motion of a kite in sky, motion of aeroplane etc.**• Displacement**

Displacement of a particle in a given time is defined as the change in the position of particle in a particular direction during that time. It is given by a vector drawn from its initial position to its final position.**• Factors Distinguishing Displacement from Distance**

—> Displacement has direction. Distance does not have direction.

—> The magnitude of displacement can be both positive and negative.

—> Distance is always positive. It never decreases with time.

—> Distance ≥ | Displacement |**• ****Uniform Speed and Uniform Velocity**

Uniform Speed. An object is said to move with uniform speed if it covers equal distances in equal intervals of time, howsoever small these intervals of time may be.

Uniform Velocity. An object is said to move with uniform velocity if it covers equal displacements in equal intervals of time, howsoever small these intervals of time may be.**• Variable Speed and Variable Velocity**

Variable Speed. An object is said to move with variable speed if it covers unequal distances in equal intervals of time, howsoever small these intervals of time may be.

Variable Velocity. An object is said to move with variable velocity if it covers unequal displacements in equal intervals of time, howsoever small these intervals of time may be.**• Average Speed and Average Velocity**

Average Speed. It is the ratio of total path length traversed and the corresponding time interval.

Or

The average speed of an object is greater than or equal to the magnitude of the average velocity over a given time interval.**• Instantaneous Speed and Instantaneous Velocity**

Instantaneous Speed. The speed of an object at an instant of time is called instantaneous speed.

Or

“Instantaneous speed is the limit of the average speed as the time interval becomes infinitesimally small”.

Instantaneous velocity

The instantaneous velocity of a particle is the velocity at any instant of time or at any point of its path.

or

“Instantaneous velocity or simply velocity is defined as the limit of the average velocity as the time interval Δt becomes infinitesimally small.”**• Acceleration**

The rate at which velocity changes is called acceleration.**• Uniform Acceleration**

If an object undergoes equal changes in velocity in equal time intervals it is called uniform acceleration.

• Average and Instantaneous Acceleration

Average Accelerating. It is the change in the velocity divided by the time-interval during which the change occurs.

Instantaneous Acceleration. It is defined as the limit of the average acceleration as the time-interval Δt goes to zero.**• Kinematical Graphs**

The ‘displacement-time’ and the ‘velocity-time’ graphs of a particle are often used to provide us with a visual representation of the motion of a particle. The ‘shape’ of the graphs depends on the initial ‘co-ordinates’ and the ‘nature’ of the acceleration of the particle (Fig.)

The following general results are always valid

(i) The slope of the displacement-time graph at any instant gives the speed of the particle at that instant.

(ii) The slope of the velocity-time graph at any instant gives the magnitude of the acceleration of the particle at that instant.

(iii) The area enclosed by the velocity-time graph, the time-axis and the two co-ordinates at ,time instants t_{1} to t_{2} gives the distance moved by the particle in the time-interval from t_{1} to t_{2}.

• Equations of Motion for Uniformly Accelerated Motion

For uniformly accelerated motion, some simple equations can be derived that relate displacement (x), time taken (f), initial velocity (u), final velocity (v) and acceleration (a). Following equation gives a relation between final and initial velocities v and u of an object moving with uniform acceleration a: v = u + at

• Suppose a body is projected vertically upward from a point A with velocity u.

In some problems it is convenient to take the downward direction as positive, in such case all the measurements in downward direction are considered as positive i.e., acceleration will be +g. But sometimes we may need to take upward as positive and if such case acceleration will be -g.**• Relative Velocity**

Relative velocity of an object A with respect to another object B is the time rate at which the object A changes its position with respect to the object B.

—> The relative velocity of two objects moving in the same direction is the difference of the speeds of the objects.

—> The relative velocity of two objects moving in opposite direction is the sum of the speeds of the objects.

•** IMPORTANT TABLES**

## CBSE Class 11 Physics Chapter-3 Important Questions –

1 Marks Questions

1. Under what condition is the relation correct?

Ans:When the particle moves with uniform velocity and along a straight line.

2.Two balls of different masses are thrown vertically upward with same initial speed. Which one will rise to a greater height?

Ans: Both the balls will rise to a greater height.

3.What is the relative velocity of two bodies having equal velocities?

Ans:If

Then

4.A railway train 400m long is going from New Delhi railway station to Kanpur. Can we consider railway train as a point object

Ans:Yes, because length of the train is smaller as compared to the distance between New Delhi and Kanpur.

5.Shipra went from her home to school 2.5km away. On finding her home closed she returned to her home immediately. What is her net displacement? What is the total distance covered by her?

Ans: Displacement = 0

Distance = 2.5km + 2.5km = 5.0km.

6.Can speed of an object be negative? Justify

Ans:No speed of an object can never be negative because distance is also always positive.

7.Under what condition the displacement and the distance of a moving object will have the same magnitude?

Ans:Distance and displacement have the same magnitude when the object moves in a straight line.

8.What is the shape of the displacement time graph for uniform linear motion?

Ans: A straight line inclined to time axis (x – axis)

9.Figure shows a displacements time graph. Comment on the sign of velocities at point P, Q, R, S and T.

Ans: Velocity at P and T is positive

Velocity at Q and S is zero

Velocity at R is negative

10. The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29:

Which of the following formulae are correct for describing the motion of the particle over the time-interval *t*2 to *t*1?

(a)

(b)

(c) *v*Average *=*

(d) *a*Average *=*

(e)

(f) *=* area under the curve bounded by the *t*-axis and the dotted line shown.

Ans. The correct formulae describing the motion of the particle are (c), (d) and, (f)

The given graph has a non-uniform slope. Hence, the formulae given in (a), (b), and (e) cannot describe the motion of the particle. Only relations given in (c), (d), and (f) are correct equations of motion.

11. In which of the following examples of motion, can the body be considered approximately a point object:

(a) a railway carriage moving without jerks between two stations.

(b) a monkey sitting on top of a man cycling smoothly on a circular track.

(c) a spinning cricket ball that turns sharply on hitting the ground.

(d) a tumbling beaker that has slipped off the edge of a table.

Ans. (a), (b)

(a) The size of a carriage is very small as compared to the distance between two stations. Therefore, the carriage can be treated as a point sized object.

(b) The size of a monkey is very small as compared to the size of a circular track. Therefore, the monkey can be considered as a point sized object on the track.

(c) The size of a spinning cricket ball is comparable to the distance through which it turns sharply on hitting the ground. Hence, the cricket ball cannot be considered as a point object.

(d) The size of a beaker is comparable to the height of the table from which it slipped. Hence, the beaker cannot be considered as a point object.

2 Marks Questions

1.Write the characteristics of displacement?

Ans: (1) It is a vector quantity having both magnitude and direction.

(2) Displacement of a given body can be positive, negative or zero.

2.Draw displacement time graph for uniformly accelerated motion. What is its shape?

Ans: The graph is parabolic in shape

3.Sameer went on his bike from Delhi to Gurgaon at a speed of 60km/hr and came back at a speed of 40km/hr. what is his average speed for entire journey.

Ans:

4.What causes variation in velocity of a particle?

Ans: Velocity of a particle changes

(1) If magnitude of velocity changes

(2) If direction of motion changes.

5.Figure. Shows displacement – time curves I and II. What conclusions do you draw from these graphs?

Ans: (1) Both the curves are representing uniform linear motion.

(2) Uniform velocity of II is more than the velocity of I because slope of curve (II) is greater.

6.Displacement of a particle is given by the expression x = 3t2 + 7t – 9, where x is in meter and t is in seconds. What is acceleration?

Ans:

7.A particle is thrown upwards. It attains a height (h) after 5 seconds and again after 9s comes back. What is the speed of the particle at a height h?

Ans:

As the particle comes to the same point as 9s where it was at 5s. The net displacement at 4s is zero.

8.Draw displacement time graph for a uniformly accelerated motion? What is its shape?

Ans: Graph is parabolic in shape

9.The displacement x of a particle moving in one dimension under the action of constant force is related to the time by the equation where x is in meters and t is in seconds. Find the velocity of the particle at (1) t = 3s (2) t = 6s.

Ans:

(i)

For

(ii) For

10.A balloon is ascending at the rate of 4.9m/s. A pocket is dropped from the balloon when situated at a height of 245m. How long does it take the packet to reach the ground? What is its final velocity?

Ans:

For packet (care of free fall) a = g = 9.8m/s2 (downwards)

Since time cannot be negative

t = 7.6s

Now

11.A car moving on a straight highway with speed of 126km/hr. is brought to stop within a distance of 200m. What is the retardation of the car (assumed uniform) and how long does it take for the car to stop?

Ans:

a = -3.06m/s2 (Retardation)

Now V = u + at

t = 11.4s

12. In Exercises 3.13 and 3.14, we have carefully distinguished between *average* speed and magnitude of *average* velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?

Ans. Instantaneous velocity is given by the first derivative of distance with respect to time i.e.,

Here, the time interval *dt* is so small that it is assumed that the particle does not change its direction of motion. As a result, both the total path length and magnitude of displacement become equal is this interval of time.

Therefore, instantaneous speed is always equal to instantaneous velocity.

13. Look at the graphs (a) to (d) (Fig. 3.20) carefully and state, with reasons, which of these *cannot* possibly represent one-dimensional motion of a particle.

(a)

(b)

(c)

(d)

Ans. (a) The given *x-t* graph, shown in (a), does not represent one-dimensional motion of the particle. This is because a particle cannot have two positions at the same instant of time.

(b) The given *v*–*t* graph, shown in (b), does not represent one-dimensional motion of the particle. This is because a particle can never have two values of velocity at the same instant of time.

(c) The given *v*–*t* graph, shown in (c), does not represent one-dimensional motion of the particle. This is because speed being a scalar quantity cannot be negative.

(d) The given *v*–*t* graph, shown in (d), does not represent one-dimensional motion of the particle. This is because the total path length travelled by the particle cannot decrease with time.

14. Figure 3.21 shows the *x-t* plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for *t* < 0 and on a parabolic path for *t* > 0? If not, suggest a suitable physical context for this graph.

Ans. No

The *x*–*t* graph of a particle moving in a straight line for *t* < 0 and on a parabolic path for *t* > 0 cannot be shown as the given graph. This is because, the given particle does not follow the trajectory of path followed by the particle as *t* = 0, *x* = 0. A physical situation that resembles the above graph is of a freely falling body held for sometime at a height.

15. A police van moving on a highway with a speed of 30 fires a bullet at a thief’s car speeding away in the same direction with a speed of. If the muzzle speed of the bullet is, with what speed does the bullet hit the thief’s car ? (Note: Obtain that speed which is relevant for damaging the thief’s car).

Ans. Speed of the police van, = 30 km/h = 8.33 m/s

Muzzle speed of the bullet, = 150 m/s

Speed of the thief’s car, = 192 km/h = 53.33 m/s

Since the bullet is fired from a moving van, its resultant speed can be obtained as:

= 150 + 8.33 = 158.33 m/s

Since both the vehicles are moving in the same direction, the velocity with which the bullet hits the thief’s car can be obtained as:

= 158.33 – 53.33 = 105 m/s

16. Figure 3.24 gives the *x-t* plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each interval.

(Fig: 3.24)

Ans. Interval 3 (Greatest), Interval 2 (Least)

Positive (Intervals 1 & 2), Negative (Interval 3)

The average speed of a particle shown in the *x*–*t* graph is obtained from the slope of the graph in a particular interval of time.

It is clear from the graph that the slope is maximum and minimum restively in intervals 3 and 2 respectively. Therefore, the average speed of the particle is the greatest in interval 3 and is the least in interval 2. The sign of average velocity is positive in both intervals 1 and 2 as the slope is positive in these intervals. However, it is negative in interval 3 because the slope is negative in this interval.

17. A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m/s. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m/s and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?

Ans. Initial velocity of the ball, *u* = 49 m/s

Acceleration, *a* =

*Case* I:

When the lift was stationary, the boy throws the ball.

Taking upward motion of the ball,

Final velocity, *v* of the ball becomes zero at the highest point.

From first equation of motion, time of ascent (*t*) is given as:

But, the time of ascent is equal to the time of descent.

Hence, the total time taken by the ball to return to the boy’s hand = 5 + 5 = 10 s.

*Case* II:

The lift was moving up with a uniform velocity of 5 m/s. In this case, the relative velocity of the ball with respect to the boy remains the same i.e., 49 m/s. Therefore, in this case also, the ball will return back to the boy’s hand after 10 s.

18. A jet airplane travelling at the speed of ejects its products of combustion at the speed of relative to the jet plane. What is the speed of the latter with respect to an observer on ground?

Ans. Speed of the jet airplane, = 500 km/h

Relative speed of its products of combustion with respect to the plane,

= – 1500 km/h

Speed of its products of combustion with respect to the ground =

Relative speed of its products of combustion with respect to the airplane,

= – 1000 km/h

The negative sign indicates that the direction of its products of combustion is opposite to the direction of motion of the jet airplane.

19. Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?

Ans. *For train A:*

Initial velocity, *u* = 72 km/h = 20 m/s

Time, *t* = 50 s

Acceleration, = 0 (Since it is moving with a uniform velocity)

From second equation of motion, distance () covered by train A can be obtained as:

= 20 × 50 + 0 = 1000 m

*For train B:*

Initial velocity, *u* = 72 km/h = 20 m/s

Acceleration, *a* =

Time, *t* = 50 s

From second equation of motion, distance () covered by train A can be obtained as:

Hence, the original distance between the driver of train A and the guard of train B is 2250 -1000 = 1250 m.

3 Marks Questions

1.Define from velocity time graph.

Ans: Slope of graph

2.A particle is moving along a straight line and its position is given by the relation

Find (a) The time at which velocity is zero.

(b) Position and displacement of the particle at that point.

(c) Acceleration for the particle at the line.

Ans:

(a)

Time cannot be negative

t = 5 seconds.

(b) Position at t = 5 s At t = 0 s

Displacement at t = 5 s and t = 0s

(c) Acceleration at t = 5s

3.A police jeep on a petrol duty on national highway was moving with a speed of 54km/hr. in the same direction. It finds a thief rushing up in a car at a rate of 126km/hr in the same direction. Police sub – inspector fired at the car of the thief with his service revolver with a muzzle speed of 100m/s. with what speed will the bullet hit the car of thief?

Ans:VPJ = 54km/hr = 15m/s VTC = 126km/hr = 35m/s

Muzzle speed of the bullet

VCP = 35 – 15 = 20m/s. VCP = Velocity of car w.r.t. police

VBC= 100 – 20 = 80 m/s VBC = Velocity of bullet w.r.t car

Thus bullet will hit the car with a velocity 80m/s.

4.Establish the relation where the letters have their usual meanings.

Ans:

Hence proved.

5.A stone is dropped from the top of a cliff and is found to ravel 44.1m diving the last second before it reaches the ground. What is the height of the cliff? g = 9.8m/s2

Ans:Let h be the height of the cliff

n be the total time taken by the stone while falling

u = 0

A = g = 9.8m/s2

Height of the cliff

h = 122.5m

6.Establish from velocity time graph for a uniform accelerated motion?

Ans:Displacement of the particle in time (t)

S = area under graph

S = area OABC

S = area of rectangle AODC + area of ADB

7.(a) Define the term relative velocity?

(b) Write the expression for relative velocity of one moving with respect to another body when objects are moving in same direction and are moving in opposite directions?

(c) A Jet airplane traveling at the speed of 500km/hr ejects its products of combustion at the speed of 1500km/h relative to the Jet plane. What is the speed of the latter with respect to an observer on the ground?

Ans: (a) Relative velocity of body A with respect to body B is defined as the time rate of change of position of A wrt. B.

(b) (i) When two objects move in the same direction

(ii) When two objects move in the opposite direction

(c) Velocity of the Jet plane VJ = 500km/hr velocity of gases wrt. Jet plane VgJ = -1500km/hr (direction is opposite)

Velocity of the Vg = -1500 + 500 = -1000km/hr

(As hot gases also comes out in opposite direction of the Jet plane)

8.Define (i) v = u + at (ii) V2 – u2 = 2as by calculus method

Ans.: We know

(i)

Integrating

Where K is constant of integration

(ii)

We know

Multiply and Divide by dx

Integrating within the limits

9. A woman starts from her home at 9.00 am, walks with a speed of on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of. Choose suitable scales and plot the *x*–*t* graph of her motion.

Ans. Speed of the woman = 5 km/h

Distance between her office and home = 2.5 km

Time taken=

It is given that she covers the same distance in the evening by an auto.

Now, speed of the auto = 25 km/h

Time taken=

The suitable *x*–*t* graph of the motion of the woman is shown in the given figure.

10. A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the *x*–*t* graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.

Ans. Distance covered with 1 step = 1 m

Time taken = 1 s

Time taken to move first 5 m forward = 5 s

Time taken to move 3 m backward = 3 s

Net distance covered = 5 – 3 = 2 m

Net time taken to cover 2 m = 8 s

Drunkard covers 2 m in 8 s.

Drunkard covered 4 m in 16 s.

Drunkard covered 6 m in 24 s.

Drunkard covered 8 m in 32 s.

In the next 5 s, the drunkard will cover a distance of 5 m and a total distance of 13 m and falls into the pit.

Net time taken by the drunkard to cover 13 m = 32 + 5 = 37 s

The *x-t* graph of the drunkard’s motion can be shown as:

11. A car moving along a straight highway with a speed of 126km is brought to a stop within a distance of 200m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?

Ans. Initial velocity of the car, u=126km/h = 35m/s

Final velocity of the car, v=0

Distance covered by the car before coming to rest, s =200m

Retardation produced in the car= a

From third equation of motion, a can be calculated as:

From first equation of motion, time (t) taken by the car to stop can be obtained as:

12. Read each statement below carefully and state with reasons and examples, if it is true or false;

A particle in one-dimensional motion

(a) with zero speed at an instant may have non-zero acceleration at that instant

(b) with zero speed may have non-zero velocity,

(c) with constant speed must have zero acceleration,

(d) with positive value of acceleration must be speeding up.

Ans. (a) True

(b) False

(c) True

(d) False

Explanation:

(a) When an object is thrown vertically up in the air, its speed becomes zero at maximum height. However, it has acceleration equal to the acceleration due to gravity (g) that acts in the downward direction at that point.

(b) Speed is the magnitude of velocity. When speed is zero, the magnitude of velocity along with the velocity is zero.

(c) A car moving on a straight highway with constant speed will have constant velocity. Since acceleration is defined as the rate of change of velocity, acceleration of the car is also zero.

(d) This statement is false in the situation when acceleration is positive and velocity is negative at the instant time taken as origin. Then, for all the time before velocity becomes zero, there is slowing down of the particle. Such a case happens when a particle is projected upwards.

This statement is true when both velocity and acceleration are positive, at the instant time taken as origin. Such a case happens when a particle is moving with positive acceleration or falling vertically downwards from a height.

13. Suggest a suitable physical situation for each of the following graphs (Fig 3.22):

(a)

(b)

(c)

Ans. (a) The given *x-t* graph shows that initially a body was at rest. Then, its velocity increases with time and attains an instantaneous constant value. The velocity then reduces to zero with an increase in time. Then, its velocity increases with time in the opposite direction and acquires a constant value. A similar physical situation arises when a football (initially kept at rest) is kicked and gets rebound from a rigid wall so that its speed gets reduced. Then, it passes from the player who has kicked it and ultimately gets stopped after sometime.

(b)In the given *v*-tgraph, the sign of velocity changes and its magnitude decreases with a passage of time. A similar situation arises when a ball is dropped on the hard floor from a height. It strikes the floor with some velocity and upon rebound, its velocity decreases by a factor. This continues till the velocity of the ball eventually becomes zero.

(c)The given *a*–*t* graph reveals that initially the body is moving with a certain uniform velocity. Its acceleration increases for a short interval of time, which again drops to zero. This indicates that the body again starts moving with the same constant velocity. A similar physical situation arises when a hammer moving with a uniform velocity strikes a nail.

14. The position-time (*x-t*) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. 3.19. Choose the correct entries in the brackets below;

(a) (A/B) lives closer to the school than (B/A)

(b) (A/B) starts from the school earlier than (B/A)

(c) (A/B) walks faster than (B/A)

(d) A and B reach home at the (same/different) time

(e) (A/B) overtakes (B/A) on the road (once/twice).

Ans. (a) A lives closer to school than B.

(b) A starts from school earlier than B.

(c) B walks faster than A.

(d) A and B reach home at the same time.

(e) B overtakes A once on the road.

Explanation:

(a) In the given *x*–*t* graph, it can be observed that distance OP < OQ. Hence, the distance of school from the A’s home is less than that from B’s home.

(b) In the given graph, it can be observed that for *x* = 0, *t* = 0 for A, whereas for *x* = 0, *t* has some finite value for B. Thus, A starts his journey from school earlier than B.

(c) In the given *x*–*t* graph, it can be observed that the slope of B is greater than that of A. Since the slope of the *x*–*t* graph gives the speed, a greater slope means that the speed of B is greater than the speed A.

(d) It is clear from the given graph that both A and B reach their respective homes at the same time.

(e) B moves later than A and his/her speed is greater than that of A. From the graph, it is clear that B overtakes A only once on the road.

4 Marks Questions

1. On a two-lane road, car A is travelling with a speed of. Two cars B and C approach car A in opposite directions with a speed of each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?

Ans. Velocity of car A, = 36 km/h = 10 m/s

Velocity of car B, = 54 km/h = 15 m/s

Velocity of car C, = 54 km/h = 15 m/s

Relative velocity of car B with respect to car A,

= 5 m/s

Relative velocity of car C with respect to car A,

= 15 + 10 = 25 m/s

At a certain instance, both cars B and C are at the same distance from car A i.e.,

*s* = 1 km = 1000 m

Time taken (*t*) by car C to cover 1000 m =

Hence, to avoid an accident, car B must cover the same distance in a maximum of 40 s.

From second equation of motion, minimum acceleration (*a*) produced by car B can be obtained as:

2. Explain clearly, with examples, the distinction between:

(a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;

(b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first.

When is the equality sign true? [For simplicity, consider one-dimensional motion only].

Ans. (a) The magnitude of displacement over an interval of time is the shortest distance (which is a straight line) between the initial and final positions of the particle.

The total path length of a particle is the actual path length covered by the particle in a given interval of time.

For example, suppose a particle moves from point A to point B and then, comes back to a point, C taking a total time *t*, as shown below. Then, the magnitude of displacement of the particle = AC.

Whereas, total path length = AB + BC

It is also important to note that the magnitude of displacement can never be greater than the total path length. However, in some cases, both quantities are equal to each other.

(b) Magnitude of average velocity=

For the given particle,

Average velocity=

Average speed=

Since (AB + BC) > AC, average speed is greater than the magnitude of average velocity. The two quantities will be equal if the particle continues to move along a straight line.

3. Figure 3.23 gives the *x-t* plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter14). Give the signs of position, velocity and acceleration variables of the particle at *t* = 0.3 s, 1.2 s, – 1.2 s.

Ans. Negative, Negative, Positive (at *t* = 0.3 s)

Positive, Positive, Negative (at *t* = 1.2 s)

Negative, Positive, Positive (at *t* = -1.2 s)

For simple harmonic motion (SHM) of a particle, acceleration (*a*) is given by the relation:

*a* =angular frequency … (i)

*t* = 0.3 s

In this time interval, *x* is negative. Thus, the slope of the *x-t* plot will also be negative. Therefore, both position and velocity are negative. However, using equation (i), acceleration of the particle will be positive.

*t* = 1.2 s

In this time interval, *x* is positive. Thus, the slope of the *x*–*t* plot will also be positive. Therefore, both position and velocity are positive. However, using equation (i), acceleration of the particle comes to be negative.

*t* = – 1.2 s

In this time interval, *x* is negative. Thus, the slope of the *x*–*t* plot will also be negative. Since both *x* and *t* are negative, the velocity comes to be positive. From equation (i), it can be inferred that the acceleration of the particle will be positive.

4. On a long horizontally moving belt (Fig. 3.26), a child runs to and fro with a speed 9 km h-1 (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4. For an observer on a stationary platform outside, what is the

(a) speed of the child running in the direction of motion of the belt ?.

(b) speed of the child running opposite to the direction of motion of the belt ?

(c) time taken by the child in (a) and (b) ?

Which of the Ans.s alter if motion is viewed by one of the parents?

(Fig: 3

Ans. (a) Speed of the belt, = 4 km/h

Speed of the boy, = 9 km/h

Since the boy is running in the same direction of the motion of the belt, his speed (as observed by the stationary observer) can be obtained as:

= 9 + 4 = 13 km/h

(b) Since the boy is running in the direction opposite to the direction of the motion of the belt, his speed (as observed by the stationary observer) can be obtained as:

= 5 km/h

(c) Distance between the child’s parents = 50 m

As both parents are standing on the moving belt, the speed of the child in either direction as observed by the parents will remain the same i.e., 9 km/h = 2.5 m/s.

Hence, the time taken by the child to move towards one of his parents is .

(d) If the motion is viewed by any one of the parents, Ans.s obtained in (a) and (b) get altered. This is because the child and his parents are standing on the same belt and hence, are equally affected by the motion of the belt. Therefore, for both parents (irrespective of the direction of motion) the speed of the child remains the same i.e., 9 km/h.

For this reason, it can be concluded that the time taken by the child to reach any one of his parents remains unaltered.

5. Two towns A and B are connected by a regular bus service with a bus leaving in either direction every *T* minutes. A man cycling with a speed of in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period *T* of the bus service and with what speed (assumed constant) do the buses ply on the road?

Ans. Let *V* be the speed of the bus running between towns A and B.

Speed of the cyclist, *v* = 20 km/h

Relative speed of the bus moving in the direction of the cyclist

= km/h

The bus went past the cyclist every 18 min i.e., (when he moves in the direction of the bus).

Distance covered by the bus =

Since one bus leaves after every *T* minutes, the distance travelled by the bus will be equal to

Both equations (i) and (ii) are equal.

Relative speed of the bus moving in the opposite direction of the cyclist

= (*V* + 20) km/h

Time taken by the bus to go past the cyclist

From equations (iii) and (iv), we get

Substituting the value of *V* in equation (iv), we get

6. Figure 3.25 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of *v* and *a* in the three intervals. What are the accelerations at the points A, B, C and D?

(Fig: 3.25)

Ans. Average acceleration is greatest in interval 2

Average speed is greatest in interval 3

*v* is positive in intervals 1, 2, and 3

*a* is positive in intervals 1 and 3 and negative in interval 2

*a* = 0 at A, B, C, D

Acceleration is given by the slope of the speed-time graph. In the given case, it is given by the slope of the speed-time graph within the given interval of time.

Since the slope of the given speed-time graph is maximum in interval 2, average acceleration will be the greatest in this interval.

Height of the curve from the time-axis gives the average speed of the particle. It is clear that the height is the greatest in interval 3. Hence, average speed of the particle is the greatest in interval 3.

*In interval 1*:

The slope of the speed-time graph is positive. Hence, acceleration is positive. Similarly, the speed of the particle is positive in this interval.

*In interval 2*:

The slope of the speed-time graph is negative. Hence, acceleration is negative in this interval. However, speed is positive because it is a scalar quantity.

*In interval 3*:

The slope of the speed-time graph is zero. Hence, acceleration is zero in this interval. However, here the particle acquires some uniform speed. It is positive in this interval.

Points A, B, C, and D are all parallel to the time-axis. Hence, the slope is zero at these points. Therefore, at points A, B, C, and D, acceleration of the particle is zero.

5 Marks Questions

1.Velocity time graph of a moving particle is shown. Find the displacement (1) 0 – 4 s (2) 0 – 8 (3) 0 12 s from the graph. Also write the differences between distance and displacement.

Ans:(1) Displacement

Diving (0 – 4) s

S1 = area of OAB s

S1 = 15 4 = 60 m

(2) Displacement (0 – 8s)

S2 = S1 + area (CDEF)

S2 = 60 + (-5) 4 = 60 -20 = 40m

(3) Displacement (0 – 12s)

S3 = S1 + area (CDEF) + area (FGHI)

S3 = 60 – 20 + 40 = 80m

Distance
| Displacement |

1. Distance is a scalar quantity
| 1. Displacement is a vector quantity. |

2. Distance is always positive | 2. Displacement can be positive negative or zero. |

2. A player throws a ball upwards with an initial speed of.

(a) What is the direction of acceleration during the upward motion of the ball?

(b) What are the velocity and acceleration of the ball at the highest point of its motion?

(c) Choose the *x* = 0 m and *t* = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of *x*-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.

(d) To what height does the ball rise and after how long does the ball return to the player’s hands? (Take *g* = 9.8 and neglect air resistance).

Ans. (a) Downward

(b) Velocity = 0, acceleration =

(c) *x* > 0 for both up and down motions, *v* < 0 for up and v > 0 for down motion, *a* > 0 throughout the motion

(d) 44.1 m, 6 s

Explanation:

(a) Irrespective of the direction of the motion of the ball, acceleration (which is actually acceleration due to gravity) always acts in the downward direction towards the centre of the Earth.

(b) At maximum height, velocity of the ball becomes zero. Acceleration due to gravity at a given place is constant and acts on the ball at all points (including the highest point) with a constant value i.e.,.

(c) During upward motion, the sign of position is positive, sign of velocity is negative, and sign of acceleration is positive. During downward motion, the signs of position, velocity, and acceleration are all positive.

(d) Initial velocity of the ball, *u* = 29.4 m/s

Final velocity of the ball, *v* = 0 (At maximum height, the velocity of the ball becomes zero)

Acceleration, *a* =

From third equation of motion, height (*s*) can be calculated as:

From first equation of motion, time of ascent (*t*) is given as:

Time of ascent = Time of descent

Hence, the total time taken by the ball to return to the player’s hands = 3 + 3 = 6 s.

3. A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between *t* = 0 to 12 s.

Ans. Ball is dropped from a height, *s* = 90 m

Initial velocity of the ball, *u* = 0

Acceleration, *a* = g =

Final velocity of the ball = *v*

From second equation of motion, time (*t*) taken by the ball to hit the ground can be obtained as:

From first equation of motion, final velocity is given as:

*v* = *u* + *at*

= 0 + 9.8 × 4.29 = 42.04 m/s

Rebound velocity of the ball, =

Time (*t*) taken by the ball to reach maximum height is obtained with the help of first equation of motion as:

0 = 37.84 + (-9.8) *t*’

Total time taken by the ball = *t* + *t*“² = 4.29 + 3.86 = 8.15 s

As the time of ascent is equal to the time of descent, the ball takes 3.86 s to strike back on the floor for the second time.

The velocity with which the ball rebounds from the floor

Total time taken by the ball for second rebound = 8.15 + 3.86 = 12.01 s

The speed-time graph of the ball is represented in the given figure as:

4. A man walks on a straight road from his home to a market 2.5 km away with a speed of 5. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h-1. What is the

(a) magnitude of average velocity, and

(b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his averages speed was zero!]

Ans. Time taken by the man to reach the market from home,

Time taken by the man to reach home from the market,

Total time taken in the whole journey = 30 + 20 = 50 min

Average velocity =

Average speed =

Time = 50 min =

Net displacement = 0

Total distance = 2.5 + 2.5 = 5 km

Average Velocity=

Average Speed=

Speed of the man = 7.5 km

Distance travelled in first 30 min = 2.5 km

Distance travelled by the man (from market to home) in the next 10 min

=

Net displacement = = 1.25 km

Total distance travelled = 2.5 + 1.25 = 3.75 km

Average velocity=

Average speed=

5. A three-wheeler starts from rest, accelerates uniformly with on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the nth second (n = 1,2,3….) versus n. What do you expect this plot to be during accelerated motion: a straight line or a parabola?

Ans. Straight line

Distance covered by a body in *n*th second is given by the relation

Where,

*u* = Initial velocity

*a =* Acceleration

*n* = Time = 1, 2, 3, ….. ,*n*

In the given case,

*u* = 0 and *a* =

This relation shows that:

… (iii)

Now, substituting different values of *n* in equation (iii), we get the following table:

| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

| 0.5 | 1.5 | 2.5 | 3.5 | 4.5 | 5.5 | 6.5 | 7.5 | 8.5 | 9.5 |

The plot between *n* and will be a straight line as shown:

Since the given three-wheeler acquires uniform velocity after 10 s, the line will be parallel to the time-axis after *n* = 10 s.

6. Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m/s and 30 m/s. Verify that the graph shown in Fig. 3.27 correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take *g* =. Give the equations for the linear and curved parts of the plot.

Ans. *For first stone*:

Initial velocity, = 15 m/s

Acceleration, *a* =

Using the relation,

Where , height of the cliff,

When this stone hits the ground, = 0

+ 15*t* + 200 = 0

40 = 0

= 0

= 0

*t* = 8 s or *t* = 5 s

Since the stone was projected at time *t* = 0, the negative sign before time is meaningless.

∴*t* = 8 s

*For second stone*:

Initial velocity, = 30 m/s

Acceleration, *a* =

Using the relation,

At the moment when this stone hits the ground; = 0

+ 30 *t* + 200 = 0

= 0

+ 4*t* + 40 = 0

= 0

(*t* + 4) = 0

*t* = 10 s or *t* = 4 s

Here again, the negative sign is meaningless.

∴*t* = 10 s

Subtracting equations (i) and (ii), we get

Equation (iii) represents the linear path of both stones. Due to this linear relation between and *t,* the path remains a straight line till 8 s.

Maximum separation between the two stones is at *t* = 8 s.

()max = 15× 8 = 120 m

This is in accordance with the given graph.

After 8 s, only second stone is in motion whose variation with time is given by the quadratic equation:

Hence, the equation of linear and curved path is given by

= 15*t* (Linear path)

(Curved path)

7. The speed-time graph of a particle moving along a fixed direction is shown in Fig. 3.28. Obtain the distance traversed by the particle between (a) *t* = 0 s to 10 s, (b) *t* = 2 s to 6 s.

(Fig. 3.28)

What is the average speed of the particle over the intervals in (a) and (b)?

Ans. (a) Distance travelled by the particle = Area under the given graph

Average speed =

(b) Let be the distances covered by the particle between time

*t* = 2 *s* to 5 *s* and *t* = 5 s to 6 s respectively.

Total distance (*s*) covered by the particle in time *t* = 2 s to 6 s

*s* = … (i)

*For distance s1:*

Let *u*’be the velocity of the particle after 2 s and *a*’ be the acceleration of the particle in *t* = 0 to *t* = 5 s.

Since the particle undergoes uniform acceleration in the interval *t* = 0 to *t* = 5 s, from first equation of motion, acceleration can be obtained as:

*v* = *u* + *at*

Where,

*v* = Final velocity of the particle

12 = 0 + *a*’ × 5

Again, from first equation of motion, we have

*v* = *u* + *at*

= 0 + 2.4 × 2 = 4.8 m/s

Distance travelled by the particle between time 2 s and 5 s i.e., in 3 s

*For distance:*

Let *a*” be the acceleration of the particle between time *t* = 5 s and *t* = 10 s.

From first equation of motion,

*v* = *u* + *at* (where *v* = 0 as the particle finally comes to rest)

0 = 12 + *a*” × 5

Distance travelled by the particle in 1s (i.e., between *t* = 5 s and *t* = 6 s)

From equations (i), (ii), and (iii), we get

*s* = 25.2 + 10.8 = 36 m

NUMERICALS

- A car is moving along x-axis. As shown in figure it moves from 0 to P in 18 seconds and return from P to Q in 6 second. What are the average velocity and average speed of the car in going from

(i) O to P (ii) o to P and back to Q

p

(i) O to P Average velocity = 20 ms-1

average speed = 20 ms-1

(ii) O to P and back to Q

Average velocity = 10 ms-1

Average speed = 20 ms-1 - On a 60km straight road, a bus travels the first 30 km with a uniform speed of 30 kmh-1. How fast must the bus travel the next 30 km so as to have average speed of 40 kmh-1 for the entire trip?

Ans. ${\text{V}}_{\text{avg}}\text{=}\frac{{\text{S}}_{\text{1}}\text{+}{\text{S}}_{\text{2}}}{{\text{t}}_{\text{1}}\text{+}{\text{t}}_{\text{2}}}\text{=}\frac{\text{S + S}}{\text{S}\left(\frac{\text{1}}{{\text{V}}_{\text{1}}}\text{+}\frac{\text{1}}{{\text{V}}_{\text{2}}}\right)}\text{=}\frac{\text{2}{\text{V}}_{\text{1}}{\text{V}}_{\text{2}}}{{\text{V}}_{\text{1}}\text{+}{\text{V}}_{\text{2}}}$“>Vavg = S1 + S2t1 + t2 = S + SS(1V1 + 1V2) = 2V1V2V1 + V2

${\text{V}}_{\text{avg}}\text{=}\frac{{\text{S}}_{\text{1}}\text{+}{\text{S}}_{\text{2}}}{{\text{t}}_{\text{1}}\text{+}{\text{t}}_{\text{2}}}\text{=}\frac{\text{S + S}}{\text{S}\left(\frac{\text{1}}{{\text{V}}_{\text{1}}}\text{+}\frac{\text{1}}{{\text{V}}_{\text{2}}}\right)}\text{=}\frac{\text{2}{\text{V}}_{\text{1}}{\text{V}}_{\text{2}}}{{\text{V}}_{\text{1}}\text{+}{\text{V}}_{\text{2}}}$ - The displacement x of a particle varies with time as x = 4t2-15t + 25. Find the position, velocity and acceleration of the particle at t = 0.
- A driver take 0.20 second to apply the breaks (reaction time). If he is driving car at a speed of 54 kmh-1 and the breaks cause a deceleration of 6.0 ms-2. Find the distance travelled by car after he sees the need to put the breaks.

Ans. (distance covered during 0.20 S) +(distance covered until rest) - A body covers 12 m in 2nd second and 20 m in 4th second. How much distance will it cover in 4 seconds after the 5th second.
A ball thrown vertically upwards with a speed of 19.6 ms-1 from the top of a tower returns to the earth in 6s. Find the height of the tower (g = 9.8 m/s2)

Two town A and B are connected by a regular bus service with a bus leaving in either direction every T min. A man cycling with a speed of 20 kmh-1 in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed do the buses ply of the road?

Ans. V=40 kmh-1 and T=9 min

1 MARK QUESTIONS

- Why can speed of a particle not be negative?

Ans. Because speed is distance travelled per second and distance is never negative. - Is it possible in straight line motion a particle have zero speed and a non-zero velocity?

Ans. No. it is not possible. - A car moving with velocity of 50 kmh-1 on a straight road is ahead of a jeep moving with velocity 75 kmh-1. How would the relative velocity be altered if jeep is ahead of car?

Ans. No change - Which of the two-linear velocity or the linear acceleration gives the direction of motion of a body?

Ans. Linear velocity - If the instantaneous velocity of a particle is zero, will Its instantaneous acceleration be necessarily zero.?

Ans. No, (highest point of vertical upward motion under gravity) - Can a body subjected to a uniform acceleration always move in the straight line?

Ans. No example. Projectile motion. - A train is moving on a straight track with acceleration a. A passenger drops a stone. What is the acceleration of stone with respect to passenger?

2 MARKS

- What are positive and negative acceleration in straight line motion?

Ans. If speed of an object increases with time, its acceleration is positive. (Acceleration is in the direction of motion) and if speed of an object decreases with time its acceleration is negative (Acceleration is opposite to the direction of motion). - Can a body have zero velocity and still be accelerating? If yes gives any situation.

Ans. Yes, at the highest point of vertical upward motion under gravity. - The displacement of a body is proportional to t3, where t is time elapsed. What is the nature of acceleration- time graph of the body?
- Suggest a suitable physical situation for the following graph.

Ans. A ball thrown up with some initial velocity rebounding from the floor with reduced speed after each hit.

- The velocity time graph for a particle is shown in figure. Draw acceleration time graph from it.

Ans. - Draw position-time graphs of two objects, A and B moving along a straight line, when their relative velocity is

(i) zero

Ans.

1 Marks Questions

1.What is “Trajectory of a projectile?

Ans:The path followed by a projectile is called trajectory of projectile e.g. parabola.

2.A projectile is fired at an angle of 30o with the horizontal with velocity 10m/s. At what angle with the vertical should it be fired to get maximum range?

Ans:Maximum range is obtained at an angle of 45o.

3. What is the value of angular speed for 1 revolution?

Ans:For one complete revolution, in time period t = T ,

4. Give an example of a body moving with uniform speed but having a variable velocity and an acceleration which remains constant in magnitude but changes in direction

Ans:A body moving in a circular path.

5.What is the direction of centripetal force when particle is following a circular path?

Ans:The direction of the centripetal force is towards the centre of the circle.

6.Two vectors are perpendicular to each other. What is the value of?

Ans: Since

7.What will be the effect on horizontal range of a projectile when its initial velocity is doubled, keeping the angle of projection same?

Ans:Four times the initial horizontal range.

8.What will be the effect on maximum height of a projectile when its angle of projection is changed from 30o to 60o, keeping the same initial velocity of projection?

Ans:Three times the initial vertical height.

9.What is the angular velocity of the hour hand of a clock?

Ans: radian per hour.

10.A body is moving on a curved path with a constant speed. What is the nature of its acceleration?

Ans:Acceleration must be perpendicular to the direction of motion and is called centripetal acceleration.

11. State, for each of the following physical quantities, if it is a scalar or a vector:

volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.

Ans. Scalar: Volume, mass, speed, density, number of moles, angular frequency

Vector: Acceleration, velocity, displacement, angular velocity

A scalar quantity is specified by its magnitude only. It does not have any direction associated with it. Volume, mass, speed, density, number of moles, and angular frequency are some of the scalar physical quantities.

A vector quantity is specified by its magnitude as well as the direction associated with it. Acceleration, velocity, displacement, and angular velocity belong to this category.

12. Pick out the two scalar quantities in the following list:

force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.

Ans. Work and current are scalar quantities.

Work done is given by the dot product of force and displacement. Since the dot product of two quantities is always a scalar, work is a scalar physical quantity.

Current is described only by its magnitude. Its direction is not taken into account. Hence, it is a scalar quantity.

13. Pick out the only vector quantity in the following list:

Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.

Ans. Impulse

Impulse is given by the product of force and time. Since force is a vector quantity, its product with time (a scalar quantity) gives a vector quantity.

1 MARK QUESTIONS

- Suggest a situation in which an object is accelerated and have constant speed

Ans. Uniform circular motion. - Two balls of different masses are thrown vertically upward with same initial velocity. Maximum heights attained by them are h1 and h2 respectively what is h1/h2?

Ans. Same height h1h2 = 1 - Will the displacement of a particle change on changing the position of origin of the coordinate system?

Ans. Will not change. - Write an example of zero vector.

Ans. The velocity vectors of a stationary object is a zero vectors. - State the essential condition for the addition of vectors.

Ans. They must represent the physical quantities of same nature.

2 Marks Questions

1.What is the angle between two forces of 2N and 3N having resultant as 4N?

Ans:Using we get

2.What is the angle of projection at which horizontal range and maximum height are equal?

Ans: Equating,

3.Prove that for elevations which exceed or fall short of 45o by equal amounts the ranges are equal?

Ans:We know

So,

R1 = R2

4.At what range will a radar set show a fighter plane flying at 3 km above its centre and at distance of 4 km from it?

Ans: Here straight distance of the object form the radar = OB

5.Two forces 5 and 10 kg wt are acting with an inclination of 120o between them. What is the angle which the resultant makes with 10kg wt?

Ans:

6.A stone is thrown vertically upwards and then it returns to the thrower. Is it a projectile? Explain?

Ans:A stone cannot be considered as a projectile because a projectile must have two perpendicular components of velocities but in this case a stone has velocity in one direction while going up or coming downwards.

7.Which is greater the angular velocity of the hour hand of a watch or angular velocity of earth around its own axis?

Ans:In hour hand of a watch (T) =1 2h

For rotation of earth T = 24h

8.Why does the direction of motion of a projectile become horizontal at the highest point of its trajectory?

Ans:At the highest point vertical component of velocity becomes zero thus direction of motion of projectile becomes horizontal.

9.A vector has magnitude 2 and another vector have magnitude 3 and is perpendicular to each other. By vector diagram find the magnitude of and show its direction in the diagram.

Ans: Here

10.Find a unit vector parallel to the resultant of the vectors

Ans: We know

11.A stone tied at the end of string is whirled in a circle. If the string breaks, the stone flies away tangentially. Why?

Ans:When a stone is moving around a circular path, its velocity acts tangent to the circle. When the string breaks, the centripetal force will not act. Due to inertia, the stone continues to move along the tangent to circular path, and flies off tangentially to the circular path.

12. What are the two angles of projection of a projectile projected with velocity 30m/s, so that the horizontal range is 45m. Take, g = 10m/s2.

Ans:

13.The blades of an aeroplane propeller are rotating at the rate of 600 revolutions per minute. Calculate its angular velocity.

Ans:

14.What is a uniform circular motion? Explain the terms time period, frequency and angular velocity. Establish relation between them.

Ans:When an object moves in a circular path with constant speed then the motion is called uniform circular motion

Time period – The time taken by the object to complete one revolution

Frequency – The total number of revolutions in one second is called the frequency.

Angular velocity – It is defined as the time rate of change of angular displacement.

15.A body of mass m is thrown with velocity at angle of 30o to the horizontal and another body B of the same mass is thrown with velocity at an angle of 60o to the horizontal. Find the ratio of the horizontal range and maximum height of A and B?

Ans:(1) When

When

(2) When

When

16. Read each statement below carefully and state with reasons, if it is true or false:

(a) The magnitude of a vector is always a scalar, (b) each component of a vector is always a scalar, (c) the total path length is always equal to the magnitude of the displacement vector of a particle. (d) the average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time, (e) Three vectors not lying in a plane can never add up to give a null vector.

Ans.

(a) True

(b) False

(c) False

(d) True

(e) True

Explanation:

(a) The magnitude of a vector is a number. Hence, it is a scalar.

(b) Each component of a vector is also a vector.

(c) Total path length is a scalar quantity, whereas displacement is a vector quantity. Hence, the total path length is always greater than the magnitude of displacement. It becomes equal to the magnitude of displacement only when a particle is moving in a straight line.

(d) It is because of the fact that the total path length is always greater than or equal to the magnitude of displacement of a particle.

(e) Three vectors, which do not lie in a plane, cannot be represented by the sides of a triangle taken in the same order.

17. State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful:

(a) adding any two scalars, (b) adding a scalar to a vector of the same dimensions, (c) multiplying any vector by any scalar, (d) multiplying any two scalars, (e) adding any two vectors, (f) adding a component of a vector to the same vector.

Ans. (a) Meaningful

(b) Not Meaningful

(c) Meaningful

(d) Meaningful

(e) Meaningful

(f) Meaningful

Explanation:

(a)The addition of two scalar quantities is meaningful only if they both represent the same physical quantity.

(b)The addition of a vector quantity with a scalar quantity is not meaningful.

(c) A scalar can be multiplied with a vector. For example, force is multiplied with time to give impulse.

(d) A scalar, irrespective of the physical quantity it represents, can be multiplied with another scalar having the same or different dimensions.

(e) The addition of two vector quantities is meaningful only if they both represent the same physical quantity.

(f) A component of a vector can be added to the same vector as they both have the same dimensions.

18. Three girls skating on a circular ice ground of radius 200 m start from a point Pon the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 4.20. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of the path skated?

Ans.

Displacement is given by the minimum distance between the initial and final positions of a particle. In the given case, all the girls start from point P and reach point Q. The magnitudes of their displacements will be equal to the diameter of the ground.

Radius of the ground = 200 m

Diameter of the ground = = 400 m

Hence, the magnitude of the displacement for each girl is 400 m. This is equal to the actual length of the path skated by girl B.

19. A man can swim with a speed of 4.0 km/h in still water. How long does he take to cross a river 1.0 km wide if the river flows steadily at 3.0 km/h and he makes his strokes normal to the river current? How far down the river does he go when he reaches the other bank?

Ans. Speed of the man, = 4 km/h

Width of the river = 1 km

Time taken to cross the river

Speed of the river, = 3 km/h

Distance covered with flow of the river =

20. A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?

Ans. Length of the string, *l* = 80 cm = 0.8 m

Number of revolutions = 14

Time taken = 25 s

Frequency,

Angular frequency,

Centripetal acceleration,

The direction of centripetal acceleration is always directed along the string, toward the centre, at all points.

21. An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.

Ans.

Radius of the loop, *r*= 1 km = 1000 m

Speed of the aircraft, *v*= 900 km/h

Centripetal acceleration,

Acceleration due to gravity, g =

22. Read each statement below carefully and state, with reasons, if it is true or false:

(a) The net acceleration of a particle in circular motion is *always* along the radius of the circle towards the centre

(b) The velocity vector of a particle at a point is *always* along the tangent to the path of the particle at that point

(c) The acceleration vector of a particle in *uniform* circular motion averaged over one cycle is a null vector

Ans.

(a) False

The net acceleration of a particle in circular motion is not always directed along the radius of the circle toward the centre. It happens only in the case of uniform circular motion.

(b) True

At a point on a circular path, a particle appears to move tangentially to the circular path. Hence, the velocity vector of the particle is always along the tangent at a point.

(c) True

In uniform circular motion (UCM), the direction of the acceleration vector points toward the centre of the circle. However, it constantly changes with time. The average of these vectors over one cycle is a null vector.

23. A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?

Ans.

Maximum horizontal distance, *R* = 100 m

The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is 45°, i.e., *θ*= 45°.

The horizontal range for a projection velocity *v*, is given by the relation:

The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity *v *is zero at the maximum height *H*.

Acceleration, *a = *Using the third equation of motion:

24. The position of a particle is given by

Where *t* is in seconds and the coefficients have the proper units for r to be in metres.

(a) Find the v and a of the particle?

(b) What is the magnitude and direction of velocity of the particle at *t* = 2.0 s?

Ans.

(a)

The position of the particle is given by:

Velocity , of the particle is given as:

Acceleration , of the particle is given as:

(b) 8.54 m/s, 69.45°below the *x*-axis

We have velocity sector,

At

The magnitude of velocity is given by:

The negative sign indicates that the direction of velocity is below the *x*-axis.

25: For any arbitrary motion in space, which of the following relations are true:

(a)

(b)

(c)

(d)

(e)

(The ‘average’ stands for average of the quantity over the time interval *t*1to *t*2)

Ans. (b)and (e)

(a)It is given that the motion of the particle is arbitrary. Therefore, the average velocity of the particle cannot be given by this equation.

(b)The arbitrary motion of the particle can be represented by this equation.

(c)The motion of the particle is arbitrary. The acceleration of the particle may also be non-uniform. Hence, this equation cannot represent the motion of the particle in space.

(d)The motion of the particle is arbitrary; acceleration of the particle may also be non-uniform. Hence, this equation cannot represent the motion of particle in space.

(e)The arbitrary motion of the particle can be represented by this equation.

26. Read each statement below carefully and state, with reasons and examples, if it is true or false:

A scalar quantity is one that

(a) is conserved in a process

(b) can never take negative values

c) must be dimensionless

(d) does not vary from one point to another in space

(e) has the same value for observers with different orientations of axes

Ans.(a) False

Despite being a scalar quantity, energy is not conserved in inelastic collisions.

(b) False

Despite being a scalar quantity, temperature can take negative values.

(c) False

Total path length is a scalar quantity. Yet it has the dimension of length.

(d) False

A scalar quantity such as gravitational potential can vary from one point to another in space.

(e) True

The value of a scalar does not vary for observers with different orientations of axes.

27. (a) A vector has magnitude and direction. Does it have a location in space? (b) Can it vary with time? (c) Will two equal vectors a and b at different locations in space necessarily have identical physical effects? Give examples in support of your answer.

Ans.(a) No; (b) Yes; (c) No

Generally speaking, a vector has no definite locations in space. This is because a vector remains invariant when displaced in such a way that its magnitude and direction remain the same. However, a position vector has a definite location in space.

A vector can vary with time. For example, the displacement vector of a particle moving with a certain velocity varies with time.

Two equal vectors located at different locations in space need not produce the same physical effect. For example, two equal forces acting on an object at different points can cause the body to rotate, but their combination cannot produce an equal turning effect.

28. (a) A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector? (b) The rotation of a body can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Does that make any rotation a vector?

Ans.(a) No; (b) No

A physical quantity having both magnitude and direction need not be considered a vector. For example, despite having magnitude and direction, current is a scalar quantity. The essential requirement for a physical quantity to be considered a vector is that it should follow the law of vector addition.

Generally speaking, the rotation of a body about an axis is not a vector quantity as it does not follow the law of vector addition. However, a rotation by a certain small angle follows the law of vector addition and is therefore considered a vector.

29. Can you associate vectors with (a) the length of a wire bent into a loop, (b) a plane area, (c) a sphere? Explain.

Ans. (a) No; (b) Yes; (c) No

(a) One cannot associate a vector with the length of a wire bent into a loop.

(b) One can associate an area vector with a plane area. The direction of this vector is normal, inward or outward to the plane area.

(c) One cannot associate a vector with the volume of a sphere. However, an area vector can be associated with the area of a sphere.

2 MARKS

- A person sitting in a train moving at constant velocity throws a ball vertically upwards. How will the ball appear to move to an observer.
- Sitting inside the train
- Standing outside the train

Ans.

(i) Vertical straight line motion

(ii) Parabolic path. - A gunman always keep his gun slightly tilted above the line of sight while shooting. Why?

Ans. Because bullet follow parabolic trajectory under constant downward acceleration. - Is the acceleration of a particle in circular motion not always towards the centre. Explain.

Ans. No, acceleration is towards the centre only in case of uniform circular motion.

3 Marks Questions

1.Derive expressions for velocity and acceleration for uniform circular motion.

OR Derive expression for linear acceleration in uniform circular motion.

Ans: (1) IF

And angular velocity

Using

Substituting in (1)

(2) Since

2.Derive an equation for the path of a projectile fired parallel to horizontal.

Ans:Let a projectile having initial uniform horizontal velocity u be under the influence of gravity, then at any instant t at position P the horizontal and vertical.

For horizontal motion

For vertical motion

We get

Or

Using equation (1) and (2)

3.(a) Define time of flight and horizontal range?

(b) From a certain height above the ground a stone A is dropped gently. Simultaneously another stone B is fired horizontally. Which of the two stones will arrive on the ground earlier?

Ans: (a) __Time of flight__ – The time taken by the projectile to complete its trajectory is called time of flight.

__Horizontal Range__ – The maximum horizontal distance covered by the projectile form the foot of the tower to the point where projectile hits the ground is called horizontal range.

(b) Both the stones will reach the ground simultaneously because the initial vertical velocity in both cases is zero and both are falling with same acceleration equal to acceleration due to gravity.

4.At what point of projectile motion (i) potential energy maximum (ii) Kinetic energy maximum (iii) total mechanical energy is maximum

Ans:(1) P.E. Will be maximum at the highest point

(P.E.) highest point = mgH

(2)K.E will be minimum at the highest point

(Vertical component of velocity is zero)

(3)Total mechanical energy

5. A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge Pof the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 4.21. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist?

Ans. (a) Displacement is given by the minimum distance between the initial and final positions of a body. In the given case, the cyclist comes to the starting point after cycling for 10 minutes. Hence, his net displacement is zero.

(b) Average velocity is given by the relation:

Average velocity

Since the net displacement of the cyclist is zero, his average velocity will also be zero.

(c) Average speed of the cyclist is given by the relation:

Average speed

Total path length = OP + PQ + QO

Time taken = 10 min

∴Average speed

6. A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity? Are the two equal?

Ans. (a) Total distance travelled = 23 km

Total time taken = 28 min

∴Average speed of the taxi

(b) Distance between the hotel and the station = 10 km = Displacement of the car

∴Average velocity

Therefore, the two physical quantities (average speed and average velocity) are not equal.

7. Rain is falling vertically with a speed of. A woman rides a bicycle with a speed of 10 in the north to south direction. What is the direction in which she should hold her umbrella?

Ans. The described situation is shown in the given figure.

Here,

= Velocity of the cyclist

= Velocity of falling rain

In order to protect herself from the rain, the woman must hold her umbrella in the direction of the relative velocity (*v*) of the rain with respect to the woman.

Hence, the woman must hold the umbrella toward the south, at an angle of nearly 18° with the vertical.

8. The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of can go without hitting the ceiling of the hall?

Ans. Speed of the ball, *u* = 40 m/s

Maximum height, *h* = 25 m

In projectile motion, the maximum height reached by a body projected at an angle *θ*, is given by the relation:

= 0.30625

sin *θ* = 0.5534

∴*θ* = (0.5534) = 33.60°

Horizontal range, *R*

9. An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30°, what is the speed of the aircraft?

Ans. The positions of the observer and the aircraft are shown in the given figure.

Height of the aircraft from ground, OR = 3400 m

Angle subtended between the positions, ∠POQ = 30

Time = 10 s

In ΔPRO:

ΔPRO is similar to ΔRQO.

∴PR = RQ

PQ = PR + RQ

= 2PR = 2 × 3400 tan 15°

= 6800 × 0.268 = 1822.4 m

∴Speed of the aircraft

10. A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the *muzzle* speed to the fixed, and neglect air resistance.

Ans. No

Range, *R* = 3 km

Angle of projection, = 30°

Acceleration due to gravity, g =

Horizontal range for the projection velocity, is given by the relation:

The maximum range (is achieved by the bullet when it is fired at an angle of 45°with the horizontal, that is,

On comparing equations (*1*) and (*2*), we get:

Hence, the bullet will not hit a target 5 km away.

4 Marks Questions

1. Given a + b + c + d = 0, which of the following statements are correct:

(a) a, b, c, and d must each be a null vector,

(b) The magnitude of (a + c) equals the magnitude of (b+ d),

(c) The magnitude of *a* can never be greater than the sum of the magnitudes of b*,* c, and d*,*

(d) b + c must lie in the plane of a and d if a and d are not collinear, and in the line

Ans. (a) Incorrect

In order to make a + b + c + d = 0, it is not necessary to have all the four given vectors to be null vectors. There are many other combinations which can give the sum zero.

(b) Correct

a + b + c + d = 0

a + c = -(b + d)

Taking modulus on both the sides, we get:

Hence, the magnitude of (a + c) is the same as the magnitude of (b + d).

(c) Correct

a + b + c + d = 0

a = (b + c + d)

Taking modulus both sides, we get:

| a | = | b + c + d |

… (*i*)

Equation (*i*) shows that the magnitude of a is equal to or less than the sum of the magnitudes of b, c, and d.

Hence, the magnitude of vector *a* can never be greater than the sum of the magnitudes of b, c, and d.

(d) Correct

For a + b + c + d = 0

a + (b + c) + d = 0

The resultant sum of the three vectors a, (b + c), and d can be zero only if (b + c) lie in a plane containing a and d, assuming that these three vectors are represented by the three sides of a triangle.

If a and d are collinear, then it implies that the vector (b + c) is in the line of a and d. This implication holds only then the vector sum of all the vectors will be zero.

2. In a harbour, wind is blowing at the speed of 72 km/h and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat?

Ans. Velocity of the boat, = 51 km/h

Velocity of the wind, = 72 km/h

The flag is fluttering in the north-east direction. It shows that the wind is blowing toward the north-east direction. When the ship begins sailing toward the north, the flag will move along

the direction of the relative velocity () of the wind with respect to the boat.

The angle between = 90° + 45°

Angle with respect to the east direction = = 0.11°

Hence, the flag will flutter almost due east.

3. A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 km/h passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit? (Take *g* =).

Ans. Height of the fighter plane = 1.5 km = 1500 m

Speed of the fighter plane, *v* = 720 km/h = 200 m/s

Let *θ* be the angle with the vertical so that the shell hits the plane. The situation is shown in the given figure.

Muzzle velocity of the gun, *u* = 600 m/s

Time taken by the shell to hit the plane = *t*

Horizontal distance travelled by the shell =

Distance travelled by the plane = *vt*

The shell hits the plane. Hence, these two distances must be equal.

= *vt*

In order to avoid being hit by the shell, the pilot must fly the plane at an altitude (*H*) higher than the maximum height achieved by the shell.

5 Marks Questions

1.(a) What is the angle between if denote the adjacent sides of a parallelogram drawn form a point and the area of the parallelogram is?

(b) State and prove triangular law of vector addition?

Ans:(a) Area of a parallelogram =

Area of parallelogram = A B Sin θ (Applying cross product)

Given, area of parallelogram =

So,

(b) Triangular law of vector addition states that if two vectors can be represented both in magnitude and direction by the sides of a triangle taken in order then their resultant is given by the third side of the triangle taken in opposite order.

Proof à in ADC

2. Establish the following vector inequalities geometrically or otherwise:

(a)

(b)

(c)

(d)

When does the equality sign above apply?

Ans.(a) Let two vectors and be represented by the adjacent sides of a parallelogram OMNP, as shown in the given figure.

Here, we can write:

In a triangle, each side is smaller than the sum of the other two sides.

Therefore, in ΔOMN, we have:

ON < (OM + MN)

If the two vectors and act along a straight line in the same direction, then we can write:

Combining equations (*iv*) and (*v*), we get:

(b) Let two vectors and be represented by the adjacent sides of a parallelogram OMNP, as shown in the given figure.

Here, we have:

In a triangle, each side is smaller than the sum of the other two sides.

Therefore, in ΔOMN, we have:

… (*iv*)

If the two vectors nd act along a straight line in the same direction, then we can write:

… (*v*)

Combining equations (*iv*) and (*v*), we get:

(c) Let two vectors and be represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.

Here we have:

In a triangle, each side is smaller than the sum of the other two sides. Therefore, in we have:

OS<OP+PS

If the two vectors act in a straight line but in opposite directions, then we can write:

Combining equations (iii) and (iv) we get:

(d) Let two vectors and be represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.

The following relations can be written for the given parallelogram.

The quantity on the LHS is always positive and that on the RHS can be positive or negative. To make both quantities positive, we take modulus on both sides as:

If the two vectors act in a straight line but in the opposite directions, then we can write:

Combining equations (iv) and (v), we get:

3. On an open ground, a motorist follows a track that turns to his left by an angle of 60° after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.

Ans.

The path followed by the motorist is a regular hexagon with side 500 m, as shown in the given figure

Let the motorist start from point P.

The motorist takes the third turn at S.

∴Magnitude of displacement = PS = PV + VS = 500 + 500 = 1000 m

Total path length = PQ + QR + RS = 500 + 500 +500 = 1500 m

The motorist takes the sixth turn at point P, which is the starting point.

∴Magnitude of displacement = 0

Total path length = PQ + QR + RS + ST + TU + UP

= 500 + 500 + 500 + 500 + 500 + 500 = 3000 m

The motorist takes the eight turn at point R

∴Magnitude of displacement = PR

Therefore, the magnitude of displacement is 866.03 m at an angle of 30° with PR.

Total path length = Circumference of the hexagon + PQ + QR

= 6 × 500 + 500 + 500 = 4000 m

The magnitude of displacement and the total path length corresponding to the required turns is shown in the given table

Turn | Magnitude of displacement (m) | Total path length (m) |

Third | 1000 | 1500 |

Sixth | 0 | 3000 |

Eighth | 866.03; 30° | 4000 |

4. A particle starts from the origin at *t* = 0 s with a velocity of and moves in the *x-y* plane with a constant acceleration of .

(a) At what time is the *x*-coordinate of the particle 16 m? What is the *y*-coordinate of the particle at that time?

(b) What is the speed of the particle at the time?

Ans.

Velocity of the particle,

Acceleration of the particle

Also,

But,

Integrating both sides:

Where,

= Velocity vector of the particle at *t* = 0

= Velocity vector of the particle at time *t*

But,

Integrating the equations with the conditions: at *t* = 0; *r* = 0 and at *t = t*; *r = r*

Since the motion of the particle is confined to the *x-y* plane, on equating the coefficients of , we get:

(a) When *x* = 16 m:

∴*y* = 10 × 2 + = 24 m

(b) Velocity of the particle is given by:

5. are unit vectors along *x*– and *y*-axis respectively. What is the magnitude and direction of the vectors, and? What are the components of a vectoralong the directions of and? [You may use graphical method]

Ans.

Consider a vector, given as:

On comparing the components on both sides, we get:

Hence, the magnitude of the vector is .

Let be the angle made by the vector, with the *x*-axis, as shown in the following figure.

Hence, the vector makes an angle of with the x-axis.

Hence, the magnitude of the vector is .

Let be the angle made by the vector, with the *x*– axis, as shown in the following figure.

Hence, the vector makes an angle of

It is given that:

On comparing the coefficients of , we have:

Let make an angle with the x-axis , as shown in the following figure.

Angle between the vectors

Component of vector , along the direction of , making an angle

Let be the angle between the vectors and .

Component of vector , along the direction of , making an angle

6. A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?

Ans. 0.86 m/s2; 54.46° with the direction of velocity

Speed of the cyclist,

Radius of the circular turn, *r* = 80 m

Centripetal acceleration is given as:

The situation is shown in the given figure:

Suppose the cyclist begins cycling from point P and moves toward point Q. At point Q, he applies the breaks and decelerates the speed of the bicycle by.

This acceleration is along the tangent at Q and opposite to the direction of motion of the cyclist.

Since the angle between is 90°, the resultant acceleration *a* is given by:

Where is the angle of the resultant with the direction of velocity

7. (a) Show that for a projectile the angle between the velocity and the *x*-axis as a function of time is given by

(b) Show that the projection angle for a projectile launched from the origin is given by

Where the symbols have their usual meaning.

Ans. (a) Let and respectively be the initial components of the velocity of the projectile along horizontal (*x*) and vertical (*y*) directions.

Let and respectively be the horizontal and vertical components of velocity at a point P.

Time taken by the projectile to reach point P = *t*

Applying the first equation of motion along the vertical and horizontal directions, we get:

(b) Maximum vertical height,

Horizontal range,

Solving equations (i) and (ii), we get: