IBDP Maths AA: Topic SL 5.9: Kinematic problems: IB style Questions HL Paper 2

Question: [Maximum mark: 7]

A particle moves in a straight line such that its velocity, vms-1, at time t seconds is given by
\(v = \frac{\left ( t^{2}+1 \right )cos t}{4}, 0\leq t\leq 3.\)
(a) Determine when the particle changes its direction of motion.
(b) Find the times when the particle’s acceleration is -1.9ms-2 
(c) Find the particle’s acceleration when its speed is at its greatest.

▶️Answer/Explanation

Ans:

(a) recognises the need to find the value of t when v = 0

\(t = 1.5707….\left ( =\frac{\pi }{2} \right )\)

\(t = 1.57\left ( =\frac{\pi }{2} \right )(s)\)

Note: Award M1A1A0 if the two correct answers are given with additional values outside 0 ≤ t ≤ 3.

(c) speed is greatest at t = 3
a = −1.8377…
a = −1.84 (m s-2)

Question: [Maximum mark: 7]

A particle moves along a straight line so that its velocity, vms-1, after t seconds is given by v(t) = esint + 4sint for 0 ≤ t ≤ 6.

(a) Find the value of t when the particle is at rest.
(b) Find the acceleration of the particle when it changes direction.
(c) Find the total distance travelled by the particle.

▶️Answer/Explanation

Ans:

(a) recognizing at rest v = 0
t = 3.34692…
t = 3.35 (seconds)
Note: Award (M1)A0 for any other solution to v = 0 eg t = −0.205 or t = 6.08.

(b) recognizing particle changes direction when v = 0 OR when t = 3.34692…

a = −4.71439…
a = −4.71(ms-2 )

(c) distance travelled = \(\int_{0}^{6}|v|dt\)  OR

= 20.7534…
= 20.8 (metres)

Question

A particle moves in a straight line. The velocity, v ms1 , of the particle at time t seconds is given by  v (t) = t sin t – 3, for 0 ≤ t ≤ 10.

The following diagram shows the graph of v .

                 

    1. Find the smallest value of t for which the particle is at rest. [2]

    2. Find the total distance travelled by the particle. [2]

    3. Find the acceleration of the particle when t = 7 . [2]

▶️Answer/Explanation

Ans:

(a) recognising  v= o  t = 6.74416… = 6.74

= 37.0968…

= 37.1 (m)

(c)

recognizing acceleration at  t = 7  is given by v′(7)

acceleration = 5.93430… 

5.93

Question

Two boats A and B travel due north.

Initially, boat B is positioned 50 metres due east of boat A.

The distances travelled by boat A and boat B, after t seconds, are x metres and y metres respectively.

The angle θ is the radian measure of the bearing of boat B from boat A.

This information is shown on the following diagram.

    1. Show that y = x + 50 cot θ.  [1]

      At time T , the following conditions are true.

      Boat B has travelled 10 metres further than boat A. Boat B is travelling at double the speed of boat A.

      The rate of change of the angle θ is 0.1 radians per second.

    2. Find the speed of boat A at time T . [6]

▶️Answer/Explanation

Ans: 

(a)

tan \(\Theta = \frac{50}{y-x}\)

OR

\(cot \Theta = \frac{y-x}{50 }\)

\(y= x+50 \: cot\Theta \)

(b)

attempt to differentiate with respect to t

\(\frac{dy}{dt}= \frac{dx}{dt}= – 50 (cosec\Theta )^{2}\frac{d\Theta }{dt}\)

attempt to set speed of B equal to double the speed of A

\(2 \frac{dx}{dt}=\frac{dx}{dt} – 50 (cosec\Theta )^{2}\frac{d\Theta }{dt}\)

\(\Theta = arctan 5 (=1.373…= 78.69..)\)

OR

\(cosec ^{2}\Theta = 1+ cot^{2}= 1+(\frac{1}{5})^2= \frac{26}{25}\)

\(\frac{dx}{dt}= -50(\frac{26}{25})\times -0.1\)

So the speed of boat A is 5.2 (ms-1)

Question

A particle moves in a straight line in a positive direction from a fixed point O.

The velocity v m \({{\text{s}}^{ – 1}}\) , at time t seconds, where \(t \geqslant 0\) , satisfies the differential equation

\[\frac{{{\text{d}}v}}{{{\text{d}}t}} = \frac{{ – v(1 + {v^2})}}{{50}}.\]

The particle starts from O with an initial velocity of 10 m \({{\text{s}}^{ – 1}}\) .

(a)     (i)     Express as a definite integral, the time taken for the particle’s velocity to decrease from 10 m \({{\text{s}}^{ – 1}}\) to 5 m \({{\text{s}}^{ – 1}}\) .

(ii)     Hence calculate the time taken for the particle’s velocity to decrease from 10 m \({{\text{s}}^{ – 1}}\) to 5 m \({{\text{s}}^{ – 1}}\) .

(b)     (i)     Show that, when \(v > 0\) , the motion of this particle can also be described by the differential equation \(\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{ – (1 + {v^2})}}{{50}}\) where x metres is the displacement from O.

(ii)     Given that v =10 when x = 0 , solve the differential equation expressing x in terms of v.

(iii)     Hence show that \(v = \frac{{10 – \tan \frac{x}{{50}}}}{{1 + 10\tan \frac{x}{{50}}}}\).

▶️Answer/Explanation

Markscheme

(a)     (i)     EITHER

Attempting to separate the variables     (M1)

\(\frac{{{\text{d}}v}}{{ – v(1 + {v^2})}} = \frac{{{\text{d}}t}}{{50}}\)     (A1)

OR

Inverting to obtain \(\frac{{{\text{d}}t}}{{{\text{d}}v}}\)     (M1)

\(\frac{{{\text{d}}t}}{{{\text{d}}v}} = \frac{{ – 50}}{{v(1 + {v^2})}}\)     (A1)

THEN

\(t = – 50\int_{10}^5 {\frac{1}{{v(1 + {v^2})}}} {\text{d}}v\,\,\,\,\,\left( { = 50\int_5^{10} {\frac{1}{{v(1 + {v^2})}}{\text{d}}v} } \right)\)     A1     N3

 

(ii)     \(t = 0.732{\text{ (sec)}}\,\,\,\,\,\left( { = 25\ln \frac{{104}}{{101}}(\sec )} \right)\)     A2     N2

[5 marks]

 

(b)     (i)     \(\frac{{{\text{d}}v}}{{{\text{d}}t}} = v\frac{{{\text{d}}v}}{{{\text{d}}x}}\)     (M1)

Must see division by v \((v > 0)\)     A1

\(\frac{{{\text{d}}v}}{{{\text{d}}x}} = \frac{{ – (1 + {v^2})}}{{50}}\)     AG     N0

 

(ii)     Either attempting to separate variables or inverting to obtain \(\frac{{{\text{d}}x}}{{{\text{d}}v}}\)     (M1)

\({\frac{{{\text{d}}v}}{{1 + {v^2}}} = – \frac{1}{{50}}\int {{\text{d}}x} }\) (or equivalent)     A1

Attempting to integrate both sides     M1

\(\arctan v = – \frac{x}{{50}} + C\)     A1A1

Note: Award A1 for a correct LHS and A1 for a correct RHS that must include C.

 

When \(x = 0{\text{ , }}v = 10{\text{ and so }}C = \arctan 10\)     M1

\(x = 50(\arctan 10 – \arctan v)\)     A1 N1

 

(iii)     Attempting to make \(\arctan v\) the subject.     M1

\(\arctan v = \arctan 10 – \frac{x}{{50}}\)     A1

\(v = \tan \left( {\arctan 10 – \frac{x}{{50}}} \right)\)     M1A1

Using tan(AB) formula to obtain the desired form.     M1

\(v = \frac{{10 – \tan \frac{x}{{50}}}}{{1 + 10\tan \frac{x}{{50}}}}\)     AG     N0

[14 marks]

Total [19 marks]

Examiners report

No comment.

Question

A body is moving through a liquid so that its acceleration can be expressed as

\[\left( { – \frac{{{v^2}}}{{200}} – 32} \right){\text{m}}{{\text{s}}^{ – 2}},\]

where \(v{\text{ m}}{{\text{s}}^{ – 1}}\) is the velocity of the body at time t seconds.

The initial velocity of the body was known to be \(40{\text{ m}}{{\text{s}}^{ – 1}}\).

(a)     Show that the time taken, T seconds, for the body to slow to \(V{\text{ m}}{{\text{s}}^{ – 1}}\) is given by

\[T = 200\int_V^{40} {\frac{1}{{{v^2} + {{80}^2}}}{\text{d}}v.} \]

(b)     (i)     Explain why acceleration can be expressed as \(v\frac{{{\text{d}}v}}{{{\text{d}}s}}\), where s is displacement, in metres, of the body at time t seconds.

  (ii)     Hence find a similar integral to that shown in part (a) for the distance, S metres, travelled as the body slows to \(V{\text{ m}}{{\text{s}}^{ – 1}}\).

(c)     Hence, using parts (a) and (b), find the distance travelled and the time taken until the body momentarily comes to rest.

▶️Answer/Explanation

Markscheme

(a)     \(\frac{{{\text{d}}v}}{{{\text{d}}t}} = – \frac{{{v^2}}}{{200}} – 32\left( { = \frac{{ – {v^2} – 6400}}{{200}}} \right)\)     (M1)

\(\int_0^T {{\text{d}}t = \int_{40}^V { – \frac{{200}}{{{v^2} + {{80}^2}}}{\text{d}}v} } \)     M1A1A1

\(T = 200\int_V^{40} {\frac{1}{{{v^2} + {{80}^2}}}{\text{d}}v} \)     AG

[4 marks]

 

(b)     (i)     \(a = \frac{{{\text{d}}v}}{{{\text{d}}t}} = \frac{{{\text{d}}v}}{{{\text{d}}s}} \times \frac{{{\text{d}}s}}{{{\text{d}}t}}\)     R1

\( = v\frac{{{\text{d}}v}}{{{\text{d}}s}}\)     AG

 

(ii)     \(v\frac{{{\text{d}}v}}{{{\text{d}}s}} = \frac{{ – {v^2} – {{80}^2}}}{{200}}\)     (M1)

\(\int_0^S {{\text{d}}s = \int_{40}^V -{\frac{{200v}}{{{v^2} + {{80}^2}}}{\text{d}}v} } \)     M1A1A1

\(\int_0^S {{\text{d}}s = \int_V^{40} {\frac{{200v}}{{{v^2} + {{80}^2}}}{\text{d}}v} } \)     M1

\(S = 200\int_V^{40} {\frac{v}{{{v^2} + {{80}^2}}}{\text{d}}v} \)     A1

[7 marks]

 

(c)     letting V = 0     (M1)

distance \( = 200\int_0^{40} {\frac{v}{{{v^2} + {{80}^2}}}{\text{d}}v = 22.3{\text{ metres}}} \)     A1

time \( = 200\int_0^{40} {\frac{1}{{{v^2} + {{80}^2}}}{\text{d}}v = 1.16{\text{ seconds}}} \)     A1

[3 marks]

 

Total [14 marks]

Examiners report

Many students failed to understand the problem as one of solving differential equations. In addition there were many problems seen in finding the end points for the definite integrals. Part (b) (i) should have been a simple point having used the chain rule, but it seemed that many students had not seen this, even though it is clearly in the syllabus.

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