**Question**

A particle moves in a straight line, starting from rest at a point O, and comes to instantaneous rest at a point P. The velocity of the particle at time ts after leaving O is v \(ms^{-2}\) , where \(\upsilon =0.6t^{2}-0.12t^{3}\)

**(i)** Show that the distance OP is 6.25 m. On another occasion, the particle also moves in the same straight line. On this occasion, the

displacement of the particle at time ts after leaving O is s m, where s=\(kt^{3}+ct^{5}\) It is given that the particle passes point P with velocity 1.25 \(m s^{-1}\) at time t = 5.**(ii)** Find the values of the constants k and c.

**(iii)** Find the acceleration of the particle at time t = 5.

**Answer/Explanation**

**(i)\( **0.6t^{2}-0.12t^{3}\)=0

(t=0 or)t=5

\(\int v dt=0.2t^{3}-0.03t^{4}

OP=[0.2\times 5^{3}-0.03\times 5^{4}]-[0]\) distance= 6.25m

** (ii)**\(k\times 5^{3}+c\times 5^{5}\)=6.25

\(v=3kt^{2}+5ct^{4}\) 125 k + 3125c = 6.25

75k+ 3125c = 1.25

k= 0.1, c = –0.002

**(iii) **a = 0.6t – 0.04\(t^{3}\)

At t = 5, a= –2 Acceleration = –2\( ms^{-2}\)

### Question

A particle P is projected vertically upwards with speed 24 m\( s^{−1}\) from a point 5 m above ground level. Find the time from projection until P reaches the ground.

**Answer/Explanation**

\(–5 = 24t – 5t ^{2}\)

\(5t ^{2}– 24t – 5 = 0 \)

t = 5

0 = 24 –\( 10t_1 → t_1\) = 2.4

0 =\( 24^2\)+ 2 × (–10) × h → h = 28.8

And 33.8 = \(\frac{1}{2}gt_{2}^{2}\rightarrow t_{2}\)=2.6

t =\( t_1 + t_2\) = 5

### Question

A particle P moves in a straight line starting from a point O. At time ts after leaving O, the displacement s m from O is given by s = \(t^{3} − 4t^{2} + 4t\) and the velocity is \(v m s^{−1}\).

(i) Find an expression for v in terms of t.

(ii) Find the two values of t for which P is at instantaneous rest.

(iii) Find the minimum velocity of P.

**Answer/Explanation**

4(i) v = \(3t ^{2}– 8t + 4\)

4(ii)\( 3t ^{2}– 8t + 4 = 0\)

t =\(\frac{2}{3}\) and t = 2

4(iii) [6t – 8 = 0]

[t =\(\frac{4}{3}, v =3( \frac{4}{3})^{2}– 8( \frac{4}{3}) + 4]\)

v = \(–\frac{4}{3}\)

[v =\( 3(t^{ 2}–\frac{8}{3}t) + 4\) = \(3(t – \frac{4}{3})^{2}+……]\)

[t =\frac{4}{3}, v = 3(t –\frac{4}{3})^{2}– \frac{4}{3}]

v = \(–\frac{4}{3}\)

### Question

A particle starts from a point O and moves in a straight line. The velocity of the particle at time ts after leaving O is v m \(s^{−1}\), where

(i) Find the acceleration of the particle during the first 5 seconds of motion.

(ii) Find the value of t when the particle is instantaneously at rest.

(iii) Find the total distance travelled by the particle in the first 10 seconds of motion.

**Answer/Explanation**

(i) Acceleration = 0.4 m\( s^{–2}\)

(ii) \(\frac{100}{t^{2}} \)– 0.1t = 0

t = 10 s

(iii) Distance t = 0 to t = 5 is

1⁄2 (1.5 + 3.5) × 5 = 12.5

\(s(t)=\int \left ( \frac{100}{t^{2}}-0.1t \right )dt\)

\(=-\frac{100}{t}-0.5t^{2}{+C}\)

\(s(10)-s(5)\)

Total distance = 12.5 + 6.25 = 18.75 m

### Question

The diagram shows the velocity-time graph of a particle which moves in a straight line. The graph consists of 5 straight line segments. The particle starts from rest at a point A at time t = 0, and initially travels towards point B on the line.

(i) Show that the acceleration of the particle between t = 3.5 and t = 6 is −10 m \(s^{−2}\).

(ii) The acceleration of the particle between t = 6 and t = 10 is 7.5 m\( s^{−2}\) . When t = 10 the velocity of the particle is V m \(s^{−1}\) . Find the value of V.

(iii) The particle comes to rest at B at time T s. Given that the total distance travelled by the particle between t = 0 and t = T is 100 m, find the value of T.

**Answer/Explanation**

(i) Acceleration =-\frac {( 25)}{2.5}\)= –10 m \(s^{–2}\)

(ii) V = –15 + 7.5 × 4

V = 15 m\( s^{–1}\)

(iii) Using v = 0 at t = 4.5 and t = 8

1⁄2 × (4.5 + 2) × 10+ 1⁄2 × (8 – 4.5) × 15 + 1⁄2 × (T – 8) × 15 = 100

T = 13.5

**Question**

A small ball is projected with speed \(15 ms^{-1}\)at an angle of 60Å above the horizontal. Find the distance

from the point of projection of the ball at the instant when it is travelling horizontally.

**Answer/Explanation**

0 = 15sin60 – gt (t = 3 3 /4 = 1.30)

x = 15cos60 x 1.3 (= 9.7428..)

y =\( (15sin60)^{2}\)/(2g) (= 8.4375..)

\(D=\sqrt{(9.74^{2}}+8.44^{2})\)

D = 12.9m

### Question

A block of mass 3 kg is initially at rest on a smooth horizontal floor. A force of 12 N, acting at an angle of \(25^{\circ}\) above the horizontal, is applied to the block. Find the distance travelled by the block in the first 5 seconds of its motion.

**Answer/Explanation**

[12 cos 25 = 3a]

a = 4 cos 25 = 3.625

[s = 1⁄2 × 4cos 25 ×\( 5^{2}\)]

Distance = 45.3 m

**Question**

A particle P moves in a straight line passing through a point O. At time ts, the velocity of P, v m\( s^{−1}\), is given by v = qt + rt2 , where q and r are constants. The particle has velocity 4 m s−1 when t = 1 and when t = 2.

(i) Show that, when t = 0.5, the acceleration of P is 4 m\( s^{−2}\).

(ii) Find the values of t when P is at instantaneous rest.

(iii) The particle is at O when t = 3. Find the distance of P from O when t = 0.

**Answer/Explanation**

(i) [q + r = 4 and 2q + 4r = 4]

q = 6 and r = –2 so v = \(6t – 2t ^{2}\)

a = 6 – 4t

At t = 0.5, a = 4

(ii) v =\( 6t – 2t ^{2}\) = 0

t = 0 and t = 3

**Question**

A particle P moves in a straight line ABCD with constant deceleration. The velocities of P at A, B and C are \(20 m s^{−1}\) , 12 \(m s^{−1}\) and \(6 m s^{−1}\) respectively.

(i) Find the ratio of distances AB : BC.

(ii) The particle comes to rest at D. Given that the distance AD is 80 m, find the distance BC

**Answer/Explanation**

(i)\([12 ^{2} = 20 ^{2} – 2a × AB\)

\(6^{2} = 12^{ 2} – 2a × BC]\)

AB = 128/a

BC = 54/a

AB : BC = 64:27

(ii)\( 0 = 20 ^{2}\) – 2a × 80 → a = 2.5

BC = 54/2.5

BC = 21.6 m

A cyclist starts from rest at a fixed point O and moves in a straight line, before coming to rest k seconds later. The acceleration of the cyclist at time ts after leaving O is am s^{−2}, where a = \(2t^{-\frac{1}{2}} – \frac{3}{5}t^{\frac{1}{2}}\) for 0 < t ≤ k.

### (a) Question

Find the value of k.

**Answer/Explanation**

Ans:

For an attempt at integration

\(v = 4t^{\frac{1}{2}} – \frac{2}{5}t^{\frac{3}{2}}\left [ +C \right ]\)

\(4t^{\frac{1}{2}} – \frac{2}{5}t^{\frac{3}{2}} = 0\)

k = 10

### (b) Question

Find the maximum speed of the cyclist.

**Answer/Explanation**

Ans:

Max speed when \(2t^{-\frac{1}{2}} – \frac{3}{5}t^{\frac{1}{2}} = 0\)

\(t = \frac{10}{3}\)

Maximum speed = 4.87 ms^{−1} to 3 sf

### (c) Question

Find an expression for the displacement from O in terms of t. Hence find the total distance travelled by the cyclist from the time at which she reaches her maximum speed until she comes to rest.

**Answer/Explanation**

Ans:

For an attempt at integration of their v

\(s = \frac{8}{3}t^{\frac{3}{2}} – \frac{4}{25}t^{\frac{5}{2}}\left [ +C \right ]\)

Substitute their t = \(\frac{10}{3}\) and t = 10

Distance = 20.7 m

### Question

A particle P is projected vertically upwards from horizontal ground with speed u m s^{−1}. P reaches a maximum height of 20 m above the ground.

(a) Find the value of u.

**Answer/Explanation**

Ans:

0 =u^{2} – 2 × 10 × 20

OR

0 =u – 10t and 20 = vt + \(\frac{1}{2}\times 10\times t^{2}\) or 20 = ut – \(\frac{1}{2}\times 10\times t^{2}\) or 20 = \(\frac{u + 0}{2}\times t\)

u = 20

### (b) Question

Find the total time for which P is at least 15m above the ground.

**Answer/Explanation**

Ans:

\(15 = 20t – \frac{1}{2}\times 10\times t^{2}\)

t = 1 or t = 3

Total time = 2 s

Alternative method for question 2(b)

\(5 = \frac{1}{2}\times 10\times t^{2}\)

t = 1

Total time = 2 s

**Question:**

A particle P moves in a straight line starting from a point O and comes to rest 14 s later. At time *t s* after leaving O, the velocity v m s^{−1} of P is given by

v = pt^{2} − qt 0 ≤ t ≤ 6,

v = 63 − 4.5t 6 ≤ t ≤ 14,

where p and q are positive constants.

The acceleration of P is zero when t = 2.

### (a) Question

Given that there are no instantaneous changes in velocity, find p and q.

**Answer/Explanation**

Ans:

a = 2 pt – q

36 p – 6p = 36

4p – q = 0

p = 3, q = 12

### (b) Question

Sketch the velocity-time graph.

**Answer/Explanation**

Ans:

Correct quadratic from t = 0 to t = 6

or

Correct straight line from 6 to 14

Both quadratic and straight line correct

All correct and key points shown

### (c) Question

Find the total distance travelled by P during the 14 s.

**Answer/Explanation**

Ans:

**Question:**

A bus moves from rest with constant acceleration for 12 s. It then moves with constant speed for 30 s before decelerating uniformly to rest in a further 6 s. The total distance travelled is 585 m.

### (a) Question

Find the constant speed of the bus.

**Answer/Explanation**

6V + 30V + 3V = 585

0.5(30 + 48)V = 585

Speed of the bus = 15 ms^{–1}

### (b) Question

Find the magnitude of the deceleration.

**Answer/Explanation**

Ans:

Magnitude of deceleration = 2.5

### Question

A particle moving in a straight line starts from rest at a point *A* and comes instantaneously to rest at a

point *B*. The acceleration of the particle at time *t *s after leaving A is am s^{−2}, where

\(a=6t^{\frac{1}{2}}-2t\)

(a) Find the value of *t* at point B. [3]

(b) Find the distance travelled from A to the point at which the acceleration of the particle is again

zero. [5]

**Answer/Explanation**

Ans

(a) \(v=\int \left ( 6t^{\frac{1}{2}}-2t \right )dt\)

\(v=4t^{\frac{3}{2}}-t^{2}(+c)\)

v = 0 leading to t = 0 or \(t^{\frac{1}{2}}=4\) leading to t = 16

(b) \(6t^{\frac{1}{2}}-2t=0\)

t = 9

\(s=\int \left ( 4t^{\frac{3}{2}}-t^{2} \right )dt\)

\([s=\frac{8}{5}t^{\frac{5}{2}}-\frac{1}{3}t^{3}(tc)]\)

\(s=\frac{8}{5}t^{\frac{5}{2}}-\frac{1}{3}t^{3}\)

Distance = 145.8m

### Question

Two cyclists, Isabella and Maria, are having a race. They both travel along a straight road with constant acceleration, starting from rest at point A. Isabella accelerates for 5 s at a constant rate am s^{−2}. She then travels at the constant speed she has reached for 10 s, before decelerating to rest at a constant rate over a period of 5 s.

Maria accelerates at a constant rate, reaching a speed of 5 m s^{−1} in a distance of 27.5 m. She then maintains this speed for a period of 10 s, before decelerating to rest at a constant rate over a period of 5 s.

(a) Given that a = 1.1, find which cyclist travels further. [5]

(b) Find the value of a for which the two cyclists travel the same distance. [2]

**Answer/Explanation**

Ans

(a) Isabella v = 5 × 1.1 = [5.5]

Use of s = ut + ½at^{2} or use of v–t graph to find total distance

\(S_{I}=\frac{1}{2}\times 1.1\times 5^{2}+10\times 5.5+\frac{1}{2}\times 1.1\times 5^{2}[=82.5]\)

or \(S_{I}=\frac{1}{2}\times (20+10)\times 5.5[=82.5]\)

\(S_{M}=27.5+5\times 10+\frac{1}{2}\times 5\times 5[=90]\)

Distances for Isabella = 82.5 and Maria = 90, so Maria goes further

(b) \(\frac{1}{2}a\times 5^{2}+10\times 5a+\frac{1}{2}a\times 5^{2}=90\)

or \(\frac{1}{2}\times(20+10)\times 5a=90 \)

a = 1.2

**Question**

A particle moves in a straight line. It starts from rest from a fixed point O on the line. Its velocity at times t s after leaving O is \(vms^{-1}\), where \(v=t^2-8t^{\frac{3}{2}}+10t\).

(a) Find the displacement of the particle from O when t=1.

(b) Show that the minimum velocity of the particle is \(-125ms^{-1}\).

**Answer/Explanation**

**Ans:**

- \([s=]\int (t^2-8t^{\frac{3}{2}}+10t)dt\)

\([s=]\frac{1}{3}t^3-\frac{16}{5}t^{\frac{5}{2}}+5t^2[+C]\)

For correct use of correct limits.

Displacement = 2.13 m (3sf) - For attempting to differentiate v.

\([a=]=2t-12t^{\frac{1}{2}}+10\)

\(a=0 \Rightarrow 2t-12t^{\frac{1}{2}}+10=0\)

\(2(t^{\frac{1}{2}}-5)(t^{\frac{1}{2}}-1)=0\) leading to t=1 or t=25

\(\frac{da}{dt}=2-6t^{\frac{1}{2}}\)

Use t = 25 in \(\frac{da}{dt}}=2-6 \times 25^{-\frac{1}{2}}\)

Evaluating \(\frac{da}{dt}\) correctly, hence a minimum.

Minimum velocity = \(25^2-8\times 25^{\frac{3}{2}}+10 \times 25 = -125 ms^{-2}\)

### Question

Two particles of masses 0.8 kg and 0.2 kg are connected by a light inextensible string that passes over

a fixed smooth pulley. The system is released from rest with both particles 0.5 m above a horizontal

floor (see diagram). In the subsequent motion the 0.2 kg particle does not reach the pulley.

(a) Show that the magnitude of the acceleration of the particles is 6 m s^{−2} and find the tension in the

string. [4]

(b) When the 0.8 kg particle reaches the floor it comes to rest.

Find the greatest height of the 0.2 kg particle above the floor. [3]

**Answer/Explanation**

(a) 0.8g – T = 0.8a, T – 0.2g = 0.2a,

For system: 0.8g – 0.2g = (0.8 + 0.2)*a*

Attempt to solve for either *a* or T

a = 6 ms^{–2} and T = 3.2 N

(b) v2 = 2 × 6 × 0.5

0 = 6 – 20s

Greatest height = 0.5 + 0.5 + 0.3 = 1.3 m

### Question

A particle P moves in a straight line. It starts from rest at a point *O* on the line and at time *t* s after

leaving *O* it has acceleration *a* m s^{−2}, where a = 6t − 18.

Find the distance *P* moves before it comes to instantaneous rest. [6]

**Answer/Explanation**

[v = 3t^{2} – 18t (+ C)]

[s = t^{3} – 9t^{2} (+ C)]

v = 3t^{2} – 18t

s = t3 – 9t^{2}

v = 0, 3t^{2} – 18t = 0 [t = 6]

s = 6^{3} – 9 × 6^{2} – [0]

s = 108 m

### Question

A particle moves in a straight line AB. The velocity vm s^{−1} of the particle t *s* after leaving A is given

by v = k(t^{2} − 10t + 21), where k is a constant. The displacement of the particle from A, in the direction

towards B, is 2.85 m when t = 3 and is 2.4 m when t = 6.

(a) Find the value of k. Hence find an expression, in terms of t, for the displacement of the particle

from A. [7]

(b) Find the displacement of the particle from A when its velocity is a minimum. [4]

**Answer/Explanation**

Ans

(a) \(\int k(t^{2}-10t+21)dt\)

\(s=k\left ( \frac{1}{3}t^{3}+5t^{2}+21t \right )+C\)

\(2.85=k\left ( \frac{1}{3}\times 3^{3}=5\times 3^{2}+21\times 3 \right )+C\ or \ 2.4=k\left ( \frac{1}{3}\times 6^{3}-5\times 6^{2}+21\times 6 \right )+C\)

2.85 = 27k + C, 2.4 = 18k + C

(A1 for both)

Solving for k

k = 0.05

\(s=0.05\left ( \frac{1}{3}t^{3}-5t^{2}+21t \right )+1.5\)

(b) Differentiating *v* or completing the square for *v *

a = 0.05(2t –10)

Min value of v is at t = 5.

Displacement at t = 5 is 2.58 m (2.5833…)

### Question

A particle *P* of mass 0.2 kg is released from rest at a point O on a plane inclined at 30° to the horizontal. At time *t s* after its release,* P* has velocity v m s^{−1 }and displacement *x* m down the plane from O. The

coefficient of friction between *P* and the plane increases as *P* moves down the plane, and equals 0.1*x*^{2}.

** (i)** Show that \(2v\frac{dv}{dx}=10-\left ( \sqrt{3} \right )x^{2}\) .[2]

** (ii)** Calculate the maximum speed of *P*. [5]

** (iii)** Find the value of *x* at the point where *P* comes to rest. [2]

**Answer/Explanation**

Ans:

### Question

A particle P is projected vertically upwards with speed 5 m s^{−1} from a point A which is 2.8 m above

horizontal ground.

(a) Find the greatest height above the ground reached by P. [3]

(b) Find the length of time for which P is at a height of more than 3.6 m above the ground. [4]

**Answer/Explanation**

Ans

(a) 0 = 5^{2} – 2gs

s = 1.25

[Height above ground =] 4.05 m

(b) Use of s = ut + ½ at^{2 }

0.8= 5t – 5t^{2}

t = 0.2 or 0.8

Length of time = 0.6 s

**Question**

A particle moves in a straight line through the point O. The displacement of the particle from O at time ts is sm, where

\(s=t^2-3t+2\) for \(0 \leq t \leq 6\),

\(s=\frac{24}{t}-\frac{t^2}{4}+25\) for \(t \geq 6\)

(a) Find the value of t when the particle is instantaneously at rest during the first 6 seconds of its motion.

At t = 6, the particle hits a barrier at a point P and rebounds.

(b) Find the velocity with which the particle arrives at P and also the velocity with which the particle leaves P.

(c) Find the total distance travelled by the particle in the first 10 second of its motion.

**Answer/Explanation**

**Ans:**

(a) [v=2t-3]

t=1.5

(b) Velocity at arrival = \(9ms^{-1}\)

\(v=-\frac{24}{t^2}-0.5t\)

Velocity when leaves = \(-3.67ms^{-1}\)

(c) At t=0, s = 2 or at t = 6, s = 20

At t = 1.5, s = -0.25

At t = 10, s = 2.4

[Total distance = 2 + 0.25 + 0.25 + 20 + (20 – 2.4)]

So total distance travelled = 40.1m

**Question**

On a straight horizontal test track, driverless vehicles (with no passengers) are being tested. A car of mass 1600 kg is towing a trailer of mass 700kg along the track. The brakes are applied, resulting in a deceleration of 12\(ms^{-2}\). The braking force acts on the car only. In addition to the braking force there are constant resistance forces of 600N on the car and of 200N on the trailer.

(a) Find the magnitude of the force in the tow-bar.

(b) Find the braking force.

(c) At the instant when the brakes are applied, the car has speed 22\(ms^{-1}\). At this instant the car is 17.5 m away from a stationary van, which is directly in front of the car.

Show that the car hits the van at a speed of \(8ms^{-1}\).

(d) After the collision, the van starts to move with speed \(5ms^{-1}\) and the car and trailer continue moving in the same direction with speed \(2ms^{-1}\).

Find the mass of the van.

**Answer/Explanation**

**Ans:**

(a) \([T-200=700 \times -12]\)

Car: \(-T-600-F=1600 \times -12\)

System: \(-600 – 200-F=2300 \times -12\)

Magnitude of T = 8200N

(b) Car \([T-F-600=1600 \times -12]\)

or

System \([-600-200-F=2300 \times -12]\)

Braking force F = 26800 N

(c) \([v^2=22^2+2 \times -12 \times 17.5]\)

\(v=8ms^{-1}\)

(d) \([2300 \times 8 + m \times 0 = 2300 \times 2 + m \times 5]\)

m = 2760 kg

**Question**

A cyclist travels along a straight road with constant acceleration. He passes through points A, B and C. The cyclist takes 2 seconds to travel each of the sections AB and BC and passes through B with speed 4.5\(ms^{-1}\). The distance AB is \(\frac{4}{5}\) of the distance BC.

(a) Find the acceleration of the cyclist.

(b) Find AC.

**Answer/Explanation**

**Ans:**

(a) Use the constant acceleration equations to obtain an expression for either \(S_{AB}\) or \(S_{BC}\) in terms of a

\(S_{AB}=2\times 4.5 – 1/2 \times a \times 2^2\)

\(S_{BC}=2 \times 4.5 + 1/2 \times a \times 2^2\)

\([2 \times 4.5 – 1/2a \times 2^2 = \frac{4}{5} (2 \times 4.5 + 1/2a \times 2^2)]\)

a = 0.5 \(ms^{-2}\)

Alternative method for question 4(a)

\([4.5=u+2a, S_{AC}=4u+8a, S_{AB}=2u + 2a]\)

Two correct equations

Three correct equations

\([2(4.5-2a)+6a=\frac{5}{4}{2(4.5-2a)+2a}]\)

\(a=0.5ms^{-2}\)

Alternative method for question 4(a)

\([AC=4.5 \times 4]\)

|\(BC=5/9 \times AC or AB = 4/9 \times AC\)

\(BC = 10 or AB = 8\)

\([10=4.5 \times 2 + 2a or 8 = 4.5 \times 2 – 2a]\)

\(A=0.5ms^{-2}\)

(b) \(S_{AB}=2 \times 4.5 – 1/2 \times 0.5 \times 2^2 = 8\)

OR

\(S_{BC}=2 \times 4.5 + 1/2 \times 0.2 \times 2^2 = 10\)

\(S_{AC}=8 + 5/4 \times 8 = 18m\)

OR

\(S_{AC}=10+4/5 \times 10 = 18 m\)

**Question**

A particle P moves in a straight line from a fixed point O. The velocity v \(ms^{-1}\) of P at time ts is given by \(v=t^{2}-8t+12\) for 0 ≤ t ≤ 8.

**(i)** Find the minimum velocity of P.

**(ii)** Find the total distance travelled by P in the interval

**Answer/Explanation**

0 ≤ t ≤ 8.

**5(i) **a = 2t – 8 a = 0→ t = 4 Minimum v = \(–4ms^{-1}\)

**5(ii) **v = 0 when (t – 2)(t – 6) = 0 t = 2 or t = 6 [s =

\(1⁄3t^{3}-4t^{2}+12t(+c)]\)

0≤ t ≤ 2 s1 = 8/3 – 16 + 24 = 32/32≤ t ≤ 6 s2 = (216/3 – 144+ 72) – (8/3 – 16+ 24)= -32/3

6≤ t ≤ 8 s3 = (512/3 – 4× \(8^{2} \)+ 12× 8) – (216/3 – 144+ 72) = 32/3 Total distance = 32 m

**Question**

A particle P is projected vertically upwards with speed \(25ms^{-1}\) from a point 3 m above horizontal ground.

**(i)** Find the time taken for P to reach its greatest height.

**(ii)** Find the length of time for which P is higher than 23 m above the ground.

**(iii)** P is higher than h m above the ground for 1 second. Find h.

**Answer/Explanation**

**2(i)** [0 = 25 – 10t] t = 2.5

**2(ii)** [20 =\( 25t – 1⁄2(gt^{2}\)] \)[t = 1 and t = 4] Required time = 4 – 1 = 3 seconds\( [v^{2}\)=\(25^{2}+ 2× (-10)× 20→v=± 15]\)[-15 = 15 – 10 T] or equivalent Required time = 1.5+ 1.5 = 3 seconds

**2(iii)** Max height reached at 2.5 s, hence reaches h after 2 s h – 3 =\( 25× 2 5\times2^{2}\) h = 33 m

Maximum height \(=1⁄2×(25+0)

× 2.5 [= 31.25] \)o.e.In 0.5 s it falls distance 1⁄2× 10 \(×0.5^{2}\)[=1.25]

h=31.25-1.25+\)=33m

**Question**

A particle moves in a straight line. It starts from rest at a fixed point O on the line. Its acceleration at time ts after leaving O is a m s^{-2}, where a = 0.4t^{3}-4.8t^{\frac{1}{2}}

**(i)** Show that, in the subsequent motion, the acceleration of the particle when it comes to instantaneous rest is 16 m s^{-2}

**(ii)** Find the displacement of the particle from O at t = 5

**Answer/Explanation**

.

**6(i) **\([\int \left ( 0.4t^{3}-4.8t^{\frac{1}{2}} \right )]\)

\(v=0.1^{4}-3.2t^{\frac{3}{2}}(+c)\)

\([v=0\rightarrow 0.1t^{4}-3.2t^{\frac{3}{2}}=0]\)

\([t^{\frac{5}{2}}=32]\)

t = 4 a = 16m s^{-2}\)

**6(ii) \(**[s=\int 0.1t^{4}-3.2t^{\frac{3}{2}}dt]

Displacement\)=\(\left [ 0.02t^{5}-1.28t^{\frac{5}{2}} \right ]_{0}^{5} \) Displacement = –9.05 m (–9.05417…)

**Question**

A particle is projected vertically upwards with speed \(30 ms^{-1}\)from a point on horizontal ground.**(i)** Show that the maximum height above the ground reached by the particle is 45 m.

**(ii)** Find the time that it takes for the particle to reach a height of 33.75 m above the ground for the

first time. Find also the speed of the particle at this time.

**Answer/Explanation**

** 2(i)** [0=30^{2}+2(-g)s]

s = maximum height = 900/20 = 45

**2(ii) **\([33.75 = 30t-1/2gt^{2}]\)

\([5r^{2}-30t+33.75=0 or 4t^{2}-24t+27=0]\)

t=1.5(reject t=4.5)

v = 30 – 1.5g = 15

** Alternative method for question 2(ii)**

v^{2}=30^{2}-2g(33.75)=225\rightarrow v=15

[33.75=1/2(30+15)\times t]

or [15=30-10t]

t = 1.5

**Question**

A particle moves in a straight line starting from rest from a point O. The acceleration of the particle

at time ts after leaving O is a \(ms^{-2}\), where

a = 5.4 − 1.62t.

(i) Find the positive value of t at which the velocity of the particle is zero, giving your answer as an exact fraction.

(ii) Find the velocity of the particle at t = 10 and sketch the velocity-time graph for the first ten seconds of the motion.

(iii) Find the total distance travelled during the first ten seconds of the motion.

**Answer/Explanation**

(i) \(v=\int (5.4-1.62t)dt

v=5.4t-0.81t^{2}(+c)

5.4-0.81t^{2}=0

t=6\frac{2}{3}=\frac{20}{3}s\)\( v=0 at t=0,negative at t= 10 and through (6\frac{2}{3},0)\)

(iii)\( s=\int (5.4t-0.81t^{2})dt

s=2.7t^{3}-0.27t^{3}(+C)

At t= 6\frac{2}{3}\),displacement =40

Total distance = 80 m

### Question

A particle P moves in a straight line. The velocity v \(m s^{−1}\) at time ts is given by

v = 4 + 0.2t for 0 ≤ t ≤ 10,

**(i)** Find the acceleration of P during the first 10 s.

**(ii)** Find the acceleration of P when t = 20.

**Answer/Explanation**

**(iii)** Sketch the velocity-time graph for 0 ≤ t ≤ 20.

**(iv)** Find the total distance travelled by P in the interval 0 ≤ t ≤ 20.

**(i)** 0.2\( (m s^{–2})\)

**(ii)** 3 1600 − a t =\ (−1600t^{3}\)

Acceleration at t = 20 is –0.2 \((m s^{–2})\)

**(iii)** Straight line joining

t = 0, v = 4 to t = 10, v = 6

Curve with correct concavity joining end of

line to t = 20, v = 0

Correct labelling on axes provided the curves

pass through (0,4), (10,6), (20,0)

**(iv)** Trapezium area = 50

\(\int (-800t^{-2})dt=-2t-800t^{-1}

\left [ -2t-800t^{-1} \right ]_{10}^{20}\)

=-40-40+20+80

Distance is 50 + 20 = 70 m

### Question

A particle is projected from a point on horizontal ground. At the instant 2 s after projection, the particle has travelled a horizontal distance of 30 m and is at its greatest height above the ground. Find the initial speed and the angle of projection of the particle. [5]

**Answer/Explanation**

Ans:

### Question

A stone is thrown with speed 9 m s^{−1 }at an angle of 60° above the horizontal from a point on horizontal ground. Find the distance between the two points at which the path of the stone makes an angle of 45° with the horizontal. [5]

**Answer/Explanation**

Ans:

### Question

A small rocket is fired vertically upwards, starting from rest at ground level, and moves with constant

acceleration. The rocket reaches a height of 200 m after 10 s.**(i)** Show that the speed of the rocket after 10 s is 40 m \(s^{−1}\) and find the acceleration of the rocket

during the first 10 s.

**(ii)** After 10 s, the rocket’s fuel stops burning and there is no upward force acting on the rocket. Find the maximum height above ground level reached by the rocket.

**(iii)** Find the total time from the instant the rocket is fired until it returns to the ground.

**Answer/Explanation**

**(i)** 200 = 1⁄2 × (0 + v) × 10

v =\( 40 m s^{–1}\)

200 = \(1⁄2 × a × 10^{2 }\)

a =\( 4 m s^{–2}\)

**(ii)** 0 =\( 40^{2}– 2 × g × s \)

s = 80 so height above ground = 280 m

**(iii)** EITHER:

0 =\( 40 – gt_{1}\)

\(t_{1}\) = 4

280 =\( 1⁄2gt_{2}^{2} \)

\(t_{2}\) = √56 = 7.48… so total time = 21.5 s

OR:

−200 = \(40t_{3} − 1⁄2gt_{3}^{2}\)

[t3 = 4 ± √56 = 4 ± 7.48]

\(t_{3}\) = 11.48 so total time is 21.5 s

**Question**

A particle P moves in a straight line starting from a point O and comes to rest 35 s later. At time ts after leaving O, the velocity v \(ms^{-1}\) of P is given by

where a and b are constants such that a > 0 and b < 0.

**(i) Show that the values of a and b are 49 and −0.04 respectively.**

**(ii)** Sketch the velocity-time graph.

**(iii)** **Find the total distance travelled by P during the 35 s.**

**Answer/Explanation**

**Question**

A racing car is moving in a straight line. The acceleration\( am s^{-1}\) at time ts after the car starts from rest is given by

where k is a constant.**(i)** Find the maximum acceleration of the car in the first five seconds of its motion. **(ii)** Find the distance of the car from its starting point when t = 5. **(iii)** The car comes to rest when t = k. Find the value of k.

**Answer/Explanation**

**(i)** [15 – 6t = 0]

Max acceleration when t = 2.5 s

Max acceleration = \(18.75ms^{-2}\)

**(ii)** [Speed =\( 7.5t^{2}-t^{3}(+c)]\)

[Distance= \(7.5t^{2}-t^{3}(+c)]\)= 2.5×125 – 0.25×625 = 156.25m

**(iii)** v(5) = 7.5×25 – 125 = 62.5\(ms^{-1}\) \(\int_{5}^{k}-\frac{625}{t^{2}}dt=\left [ \frac{625}{t} \right ]_{5}^{k}

=\frac{625}{k}-\frac{625}{5}=\frac{625}{k}-125

\frac{625}{k}-125=v(k)-v(5)=-62.5\) k=10

\(v(5)=7.5\times 125=62.5ms^{-1}

v(t)=\int -\frac{625}{t^{2}}dt=\frac{625}{t}+c

[c=-62.5]

v(t)=\frac{625}{t}-62.5

v(k)=\frac{625}{k}-62.5

=0\) k=10

**Question**

A particle P is projected vertically upwards from a point O. When the particle is at a height of 0.5 m, its speed is\( 6 ms^{-1}\) . Find

**(i)** the greatest height reached by the particle above O, **(ii)** the time after projection at which the particle returns to O.

**Answer/Explanation**

**(i)** [0 =\( 6^{2} – 2g × s]\)

s = 1.8 Total height = 2.3m

\([6^{2}=u^{2}-2g\times 0.5]

u^{2}=46

0^{2}=46-2gs\rightarrow s=total height=2.3m\)

**(ii)** [2.3 = 0 \(+0.5gt^{2}\)]

t = 0.678

Total time = 2 × 0.678 = 1.36 s

[ 0 46 =\(\sqrt{46} \)− gt ]

\(t=\frac{\sqrt{46}}{10}=0.678\)

Total time = 2 × 0.678 = 1.36 s

**Question**

A particle of mass 0.1 kg is released from rest on a rough plane inclined at 20Å to the horizontal. It is given that, 5 seconds after release, the particle has a speed of \(2 ms^{-1}\)

**(i)** Find the acceleration of the particle and hence show that the magnitude of the frictional force

acting on the particle is 0.302 N, correct to 3 significant figures. **(ii)** Find the coefficient of friction between the particle and the plane.

**Answer/Explanation**

**(i)** 2 = 5a → a =\( 0.4ms^{-2}\)

[0.1g sin20 – F = 0.1 × 0.4]

F = 0.302N

**(ii)** [R = 0.1g cos20 ( = 0.9397)]

μ = 0.3020/0.9397 = 0.321

### Question

A particle *P* moves in a straight line. The velocity *v* m s^{−1} at time *t s* is given by

* v =* 5*t*(*t* − 2) for 0 ≤ *t* ≤ 4,* v = k* for 4 ≤ *t *≤ 14,* v =* 68 − 2*t* for 14 ≤ *t* ≤ 20

where *k* is a constant.

* ***(i)** Find k. [1]

* ***(ii)** Sketch the velocity-time graph for 0 ≤ t ≤ 20. [3]

* ***(iii)** Find the set of values of t for which the acceleration of P is positive. [2]

* ***(iv)** Find the total distance travelled by P in the interval 0 ≤ t ≤ 20. [5]

**Answer/Explanation**

### Question

A uniform lamina is made by joining a rectangle *ABCD*, in which *AB = CD =* 0.56 m and *BC = AD* = 2 m, and a square *EFGA* of side 1.2 m. The vertex *E* of the square lies on the edge *AD* of the rectangle (see diagram). The centre of mass of the lamina is a distance *h* m from *BC* and a distance *v* m from *BAG*.

**(i)** Find the value of *h* and show that *v = h*. [4]

The lamina is freely suspended at the point *B* and hangs in equilibrium.

**(ii)** State the angle which the edge *BC* makes with the horizontal. [1]

Instead, the lamina is now freely suspended at the point *F* and hangs in equilibrium.

**(iii)** Calculate the angle between *FG* and the vertical. [2]

**Answer/Explanation**

Ans:

### Question

Two particles *A *and *B*, of masses 0.8 kg and 0.2 kg respectively, are connected by a light inextensible string. Particle *A* is placed on a horizontal surface. The string passes over a small smooth pulley *P *fixed at the edge of the surface, and *B* hangs freely. The horizontal section of the string, *AP*, is of length 2.5 m. The particles are released from rest with both sections of the string taut.

** (i)** Given that the surface is smooth, find the time taken for *A* to reach the pulley. [5]

** (ii)** Given instead that the surface is rough and the coefficient of friction between *A* and the surface is 0.1, find the speed of *A* immediately before it reaches the pulley. [5]

**Answer/Explanation**

Ans:

**First Alternative Method for (i) **

**Second Alternative Method for (i)**

**First Alternative Method for (ii)**

**Second Alternative Method for (ii)**

### Question

A car of mass 1200 kg is pulling a trailer of mass 800 kg up a hill inclined at an angle ! to the horizontal, where sin α = 0.1. The system of the car and the trailer is modelled as two particles connected by a light inextensible cable. The driving force of the car’s engine is 2500 N and the resistances to the car and trailer are 100 N and 150 N respectively.

** (i)** Find the acceleration of the system and the tension in the cable. [4]

** (ii)** When the car and trailer are travelling at a speed of 30 m s^{−1}, the driving force becomes zero. The cable remains taut. Find the time, in seconds, before the system comes to rest. [3]

**Answer/Explanation**

Ans:

**(i)** [2500 – 2000*g* × 0.1 – 250 = 2000*a*]

* a* = 1/8 = 0.125ms^{–2}

* *2500 – *T* – 100 – 1200*g* × 0.1 = 1200 × 0.125* *or* T* – 150 – 800*g* × 0.1 = 800 × 0.125

* T* = 1050N

**(ii) **–2000g × 0.1 – 250 = 2000*a** *[a = – 1.125]

* *0 = 30 – 1.125*t *

* t* = 26.7 s

**Alternative method for 5(ii)**

**(ii) **[½ (2000) 30^{2 }= 250s + 2000 × *g* × 0.1*s*]

→ *s* = 400

* t* = 26.7 s

### Question

A cyclist starts from rest at point *A* and moves in a straight line with acceleration 0.5 m s^{−2 }for a distance of 36 m. The cyclist then travels at constant speed for 25 s before slowing down, with constant deceleration, to come to rest at point *B*. The distance *AB* is 210 m.

** (i)** Find the total time that the cyclist takes to travel from *A *to* B*. [5]

24 s after the cyclist leaves point **A**, a car starts from rest from point **A**, with constant acceleration 4 m s^{−2}, towards *B*. It is given that the car overtakes the cyclist while the cyclist is moving with constant speed.

** (ii)** Find the time that it takes from when the cyclist starts until the car overtakes her. [5]

**Answer/Explanation**

**(i)** 36 = 0 + 0.5 × 0.5t^{2}

t = 12

v^{2 }= 0 + 2 × 0.5 × 36

v = 6

s = 6 × 25

remaining distance

= 210 – 36 – 150 = 24

24 = (6 + 0)/ 2 × t

t = 8

Total Time = 12 + 25 + 8 = 45 s

** (ii) **Distance travelled by cyclist

= 36 + 6(t – 12)

Distance travelled by car

= 0.5 × 4 × (t – 24)^{2}

2t^{2} – 96t + 1152

= 36 + 6t – 72

[t^{2}– 51t + 594 = 0]

t = 33 or t = 18

Time = 33 s

### Question

A particle *P* moves in a straight line, starting from a point *O*. The velocity of *P*, measured in m s^{−1}, at time *t s* after leaving *O* is given by

v = 0.6t − 0.03t^{2}.

** (i)** Verify that, when t = 5, the particle is 6.25 m from O. Find the acceleration of the particle at this time. [4]

** (ii)** Find the values of t at which the particle is travelling at half of its maximum velocity. [6]

**Answer/Explanation**

Ans:

** (i)** s = 0.3t^{2} – 0.01t^{3}

s(5) = 0.3 × 5^{2} – 0.01 × 5^{3} = 6.25

a = 0.6 – 0.06t

a(5) = 0.6 – 0.0 × 5 = 0.3 ms^{-2}

** (ii) **Maximum velocity is when

0.6 – 0.06t = 0

[*t* = 10]

Max velocity = 3 ms^{–1}

0.6t – 0.03t^{2}= 1.5

[t2– 20t + 50 = 0]

Times are 2.93 s

and 17.07 s

### Question

A particle of mass 0.5 kg starts from rest and slides down a line of greatest slope of a smooth plane. The plane is inclined at an angle of 30° to the horizontal.

** (i)** Find the time taken for the particle to reach a speed of 2.5 m s^{−1}. [3]

** **When the particle has travelled 3 m down the slope from its starting point, it reaches rough horizontal ground at the bottom of the slope. The frictional force acting on the particle is 1 N.

** (ii)** Find the distance that the particle travels along the ground before it comes to rest. [3]

**Answer/Explanation**

Ans:

**2 (i)** *a* = *g*sin30 = 5

2.5 = 0 + 5*t ** t* = 0.5 Time = 0.5 s

**(ii)** v^{2}= 0 + 2 × 5 × 3 = 30 * *–1 = 0.5*a* → *a* = –2* *0 = 30 + 2 × (–2) × *s* * *Distance = 7.5m

**First alternative method for 2(ii)**

* *v^{2}= 0 + 2 × 5 × 3 = 30 * *0.5 × 0.5 × 30 = 1 × distance * *Distance = 7.5m

**Second alternative method for 2(ii)**

* *PE lost = 0.5 × 10 × 3 sin30 = 7.5* *7.5 = 1 × distance * *Distance = 7.5m

### Question

Two particles A and B start to move at the same instant from a point O. The particles move in the same direction along the same straight line. The acceleration of A at time t s after starting to move is am s^{−2 }, where a = 0.05 − 0.0002t.

** (i)** Find A’s velocity when t = 200 and when t = 500. [4]

** **** **B moves with constant acceleration for the first 200 s and has the same velocity as A when t = 200. B moves with constant retardation from t = 200 to t = 500 and has the same velocity as A when t = 500.

** (ii)** Find the distance between A and B when t = 500. [6]

**Answer/Explanation**

Ans:

** (i)** v(t) = 0.05t – 0.0001t^{2}(+ 0)

v(200) = 10 – 4 = 6 ms^{–1}

v(500) = 25 – 25 = 0

** (ii)**

**Question**

A particle of mass 3 kg falls from rest at a point 5 m above the surface of a liquid which is in a container. There is no instantaneous change in speed of the particle as it enters the liquid. The depth of the liquid in the container is 4 m. The downward acceleration of the particle while it is moving in the liquid is \(5.5 ms^{-2}\)

.**(i)** Find the resistance to motion of the particle while it is moving in the liquid.**(ii)** Sketch the velocity-time graph for the motion of the particle, from the time it starts to move until the time it reaches the bottom of the container. Show on your sketch the velocity and the time when the particle enters the liquid, and when the particle reaches the bottom of the container.

**Answer/Explanation**

**(i)** [3g – R = 3 × 5.5]

Resistance is 13.5N **(ii)** Graph consists of two line segments; the first starts at the origin and has a positive gradient. The second starts where first one ends and has positive but less steep gradient.

[\(vs^{2}\) = 2 × 10 × 5 = 100 or

\(VB^{2}\)=\(vt^{2}\)+ 2 × 5.5 × 4]

\(v_s = 10 ms^{-1}\) at surface and

\(v_ b= 12 ms^{-1}\) at bottom

\(10 = 0 + 10t_1 \) or

\(12 = 10 + 5.5(t_2 – t_1)\)

For using v = u + at

\(t_1 = 1 s \) at surface and shown on sketch

\(t_2 = 1.36 s \) at bottom and shown on sketch.

**Question**

Particles P and Q move on a straight line AOB. The particles leave O simultaneously, with P moving towards A and with Q moving towards B. The initial speed of P is 1.3 \(ms^{-1}\) and its acceleration in the direction OA is 0.1 \(ms^{-2}\) . Q moves with acceleration in the direction OB of 0.016t \(ms^{-2}\) , where t seconds is the time elapsed since the instant that P and Q started to move from O. When t = 20, particle P passes through A and particle Q passes through B.**(i)** Given that the speed of Q at B is the same as the speed of P at A, find the speed of Q at time t = 0. [4]**(ii)** Find the distance AB.

**Answer/Explanation**

**(i)** End speed = 1.3 + 0.1 × 20

\(v_{Q}(t)=0.008t^{2}+v_{Q}(0)\)

\([3.3=0.008t^{2}\times 20^{2}+v_{Q}(0)]\)

Speed of Q when t = 0 is 0.1 \(ms^{-1}\)

**(ii) **Distance AO \(= (3.32^2– 1.32^2) ÷ (2 × 0.1)\) or

\(20^2 × 1⁄2 (1.3 + 3.3) [ = 46]\)

or AO\( = 1.3(20)^3 + 1⁄2(0.1) × 20^2\)

Distance OB \(= 0.008 × 203 ÷ 3 + 0.1 × 20\)

\(= \frac{70}{3} = 23.3\) Distance AB is 69.3 m

**Question**

Two cyclists P and Q travel along a straight road ABC, starting simultaneously at A and arriving simultaneously at C. Both cyclists pass through B 400 s after leaving A. Cyclist P starts with speed \(3ms^{-1}\)and increases this speed with constant acceleration 0.005 \(m s^{-2} \)until he reaches B.**(i)** Show that the distance AB is 1600 m and find P’s speed at B. Cyclist Q travels from A to B with speed \(v ms^{-1}\) at time t seconds after leaving A, where

v = 0.04t −\( 0.0001t^{2} \)+k, and k is a constant.

**(ii)** Find the value of k and the maximum speed of Q before he has reached B. Cyclist P travels from B to C, a distance of 1400 m, at the speed he had reached at B. Cyclist Q travels from B to C with constant acceleration a\(m s^{-1}\)

.**(iii)** Find the time taken for the cyclists to travel from B to C and find the value of a.

**Answer/Explanation**

**(i) **AB = 3 × 400

\(\frac{1}{2}0.005\times 400^{2}\)=1600m

(AG)**or** \(v_{B}==3+0.005\times 400\)\(=5ms^{-1} v_{B}==3+0.005\times 400=5ms^{-1}\)**or**

AB= \(3\times 400+\frac{1}{2}0.005\times 400^{2}\)=1600m (AG)**(ii)** \([0.02t^{2}-0.0001t^{3}/3+kt]_{0}^{400}\)=1600

400k = 1600 – 0.02 × \(400^{2}\) + 0.0001 ×\( 400^{3}\) ÷ 3 →k = 4 – 8 + 16/3 = 4/3

[dv/dt = 0.04 – 0.0002t

(= 0 when t = 200)

vmax = 0.04 × 200 – 0.0001 ×\( 200^{2}\)+ 4/3

Maximum speed is\( 5.33 ms^{-1}\)

**(iii) ** Time taken is 280s

[1400 = 4/3 × 280 + \(\frac{1}{2}280^{2}a]\)

a = 0.0262

**Question**

Particles A of mass 0.25 kg and B of mass 0.75 kg are attached to opposite ends of a light inextensible string which passes over a fixed smooth pulley. The system is held at rest with the string taut and its straight parts vertical. Both particles are at a height of h m above the floor (see Fig. 1). The system is released from rest, and 0.6 s later, when both particles are in motion, the string breaks. The particle A does not reach the pulley in the subsequent motion.**(i)** Find the acceleration of A and the distance travelled by A before the string breaks.

The velocity-time graph shown in Fig. 2 is for the motion of particle A until it hits the floor. The velocity of A when the string breaks is\( V ms^{-1}\) and T s is the time taken for A to reach its greatest height.**(ii)** Find the value of V and the value of T. **(iii)** Find the distance travelled by A upwards and the distance travelled by A downwards and hence find h.

**Answer/Explanation**

**(i) **Acceleration is\( 5ms^{-2}\) Distance is 0.9m

**(ii)** \( \frac{1}{2}0.6\)× V = 0.9 → V = 3 T = 0.9

\([S_{up}=\frac{1}{2}0.9\times 3 and ]

S_{down}=0+\frac{1}{2}g(1.6-0.9)^{2}]\)

Distance upwards is 1.35m and distance downwards is 2.45m h = 1.1

**Question**

A particle is projected vertically upwards with speed 9 m s−1 from a point 3.15 m above horizontal ground. The particle moves freely under gravity until it hits the ground. For the particle’s motion from the instant of projection until the particle hits the ground, find the total distance travelled and the total time taken.

**Answer/Explanation**

For s = 4.05

Total distance = 4.05 + (3.15 + 4.05)

= 11.25m

tupwards = 0.9

For downwards motion

(3.15 + 4.05) = \(\frac{1}{2}gt^{2}\rightarrow t=1.2\)

Time taken is 2.1s

Alternative Mark Scheme for final 3 marks

\([–3.15 = 9T +\frac{1}{2}(-g)T^{2}]\)

\([100t^{2}-180t-63=0]\)

(10T – 21)(10T + 3) = 0

### Question

A particle P starts from rest at a point O and moves in a straight line. P has acceleration 0.6t m s^{−2 }at time t seconds after leaving O, until t = 10.

** (i)** Find the velocity and displacement from O of P when t = 10. [5]

** **After t = 10, P has acceleration −0.4t m s^{−2} until it comes to rest at a point A.

** (ii)** Find the distance OA. [7]

**Answer/Explanation**

Ans:

### Question

Particles A and B, of masses 0.3 kg and 0.7 kg respectively, are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley. A is held at rest and B hangs freely, with both straight parts of the string vertical and both particles at a height of 0.52 m above the floor (see diagram). A is released and both particles start to move.

** (i)** Find the tension in the string. [4]

** **When both particles are moving with speed 1.6 m s^{−1} the string breaks.

** (ii)** Find the time taken, from the instant that the string breaks, for A to reach the floor. [5]

**Answer/Explanation**

**Alternative Marking Scheme for the last three marks **

### Question

A cyclist exerts a constant driving force of magnitude F N while moving up a straight hill inclined at an angle α to the horizontal, where sin \(\alpha =\frac{36}{325}\) . A constant resistance to motion of 32 N acts on

the cyclist. The total weight of the cyclist and his bicycle is 780 N. The cyclist’s acceleration is −0.2 m s^{−2}.

** (i)** Find the value of F. [4]

** **The cyclist’s speed is 7 m s^{−1} at the bottom of the hill.

** (ii)** Find how far up the hill the cyclist travels before coming to rest. [2]

**Answer/Explanation**

**(i) **F – 780 × (36÷325) – 32

= 78 × (–0.2)

F = 103 (102.8 exact)

** (ii)** [0 = 7^{2}+ 2(–0.2)s]

Distance is 122.5m

(accept 122 or 123)

### Question

A car driver makes a journey in a straight line from A to B, starting from rest. The speed of the car increases to a maximum, then decreases until the car is at rest at B. The distance travelled by the car t seconds after leaving A is 0.000 011 7(400t^{3} − 3t^{4}) metres.

**(i)** Find the distance AB. [3]

**(ii)** Find the maximum speed of the car. [4]

**(iii)** Find the acceleration of the car

** (a)** as it starts from A,

** (b)** as it arrives at B. [2]

**(iv)** Sketch the velocity-time graph for the journey. [2]

**Answer/Explanation**

### Question

A train of mass 400 000 kg is moving on a straight horizontal track. The power of the engine is constant and equal to 1500 kW and the resistance to the train’s motion is 30 000 N. Find

**(i)** the acceleration of the train when its speed is 37.5 m s^{−1}, [4]

**(ii)** the steady speed at which the train can move. [2]

**Answer/Explanation**

Ans:

**(i)** DF = 1500 000/37.5 (= 40 000)

[DF – R = ma]

DF – 30 000 = 400 000a

Acceleration is 0.025 ms–2

** (ii)** [1500 000/v – 30 000 = 0]

Steady speed is 50 ms–1

### Question

The top of a cliff is 40 metres above the level of the sea. A man in a boat, close to the bottom of the cliff, is in difficulty and fires a distress signal vertically upwards from sea level. Find

**(i)** the speed of projection of the signal given that it reaches a height of 5 m above the top of the cliff, [2]

**(ii)** the length of time for which the signal is above the level of the top of the cliff. [2]

The man fires another distress signal vertically upwards from sea level. This signal is above the level of the top of the cliff for √(17) s.

**(iii)** Find the speed of projection of the second signal. [3]

**Answer/Explanation**

Ans:

**Special Ruling** for candidates who assume, without justification, that the length of time required is that of the upward movement only. (maximum mark 1).

**Question**

Particle P travels along a straight line from A to B with constant acceleration 0.05 m\( s^{−2}\) . Its speed at A is 2 m\( s^{−1}\) and its speed at B is 5 m \(s^{−1}\).

**(i)** Find the time taken for P to travel from A to B, and find also the distance AB. **Particle Q also travels along the same straight line from A to B, starting from rest at A. At time ts after leaving A, the speed of Q is kt3 m\( s^{−1}\) , where k is a constant. Q takes the same time to travel from A to B as P does.****(ii) Find the value of k and find Q’s speed at B.**

**Answer/Explanation**

**(i)** [5 = 2 + 0.05t or 25 = 4 + 2 × 0.05(AB)]

Time taken is 60 s (or Distance is 210 m)

Distance is 210 m (or Time taken is 60 s)

**(ii)** s =\( kt^{4}\/4 (+C)

C = 0 (may be implied by its absence)

[210 = k × \(60^{4}\)/4]

k = 7/108000 or 0.0000648

Speed of Q at B is 14 \(ms^{-1}\)

**Question**

Particle P travels along a straight line from A to B with constant acceleration 0.05 m\( s^{−2}\) . Its speed at A is 2 m\( s^{−1}\) and its speed at B is 5 m \(s^{−1}\). ** (i) Find the time taken for P to travel from A to B, and find also the distance AB. **

Particle Q also travels along the same straight line from A to B, starting from rest at A. At time ts after leaving A, the speed of Q is kt3 m\( s^{−1}\) , where k is a constant. Q takes the same time to travel from A to B as P does.**(ii)** Find the value of k and find Q’s speed at B.

**Answer/Explanation**

**(i)** [5 = 2 + 0.05t or 25 = 4 + 2 × 0.05(AB)]

Time taken is 60 s (or Distance is 210 m)

Distance is 210 m (or Time taken is 60 s)

**(ii)** s =\( kt^{4}\/4 (+C)

C = 0 (may be implied by its absence)

[210 = k × \(60^{4}\)/4]

k = 7/108000 or 0.0000648

Speed of Q at B is 14 \(ms^{-1}\)

**Question**

A particle P starts at the point O and travels in a straight line. At time t seconds after leaving O the

velocity of P is vm\( s^{−1}\), where v = 0.75t2 − \(0.0625t ^{3}\) . Find

**(i)** the positive value of t for which the acceleration is zero, **(ii)** the distance travelled by P before it changes its direction of motion

**Answer/Explanation**

.

**(i)** [a =\( 1.5t – 0.1875{t^2}]\)

[0.1875t(8 – t) = 0]

Acceleration is zero when t = 8

**(ii)** Changes direction when t = 12

s =\( 0.25t^{3}\) – \(0.0625t^{4}\) ÷ 4 (+ C)

[s = 0.25 × 1728 – 0.0625 × 20736 ÷ 4]

Distance is 108 m