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Date	May 2021	Marks available	3	Reference code	21M.1.AHL.TZ2.11
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 2
Command term	Show that and Hence	Question number	11	Adapted from	N/A

Question

The acceleration, $a {ms}^{- 2}$ , of a particle moving in a horizontal line at time $t$ seconds, $t \geq 0$ , is given by $a = - (1 + v)$ where $v {ms}^{- 1}$ is the particle’s velocity and $v > - 1$ .

At $t = 0$ , the particle is at a fixed origin $O$ and has initial velocity $v_{0} {ms}^{- 1}$ .

Initially at $O$ , the particle moves in the positive direction until it reaches its maximum displacement from $O$ . The particle then returns to $O$ .

Let $s$ metres represent the particle’s displacement from $O$ and $s_{max}$ its maximum displacement from $O$ .

Let $v (T - k)$ represent the particle’s velocity $k$ seconds before it reaches $s_{max}$ , where

$v (T - k) = (1 + v_{0}) e^{- (T - k)} - 1$ .

Similarly, let $v (T + k)$ represent the particle’s velocity $k$ seconds after it reaches $s_{max}$ .

By solving an appropriate differential equation, show that the particle’s velocity at time $t$ is given by $v (t) = (1 + v_{0}) e^{- t} - 1$ .

[6]

Show that the time $T$ taken for the particle to reach $s_{max}$ satisfies the equation $e^{T} = 1 + v_{0}$ .

[2]

b.i.

By solving an appropriate differential equation and using the result from part (b) (i), find an expression for $s_{max}$ in terms of $v_{0}$ .

[5]

b.ii.

By using the result to part (b) (i), show that $v (T - k) = e^{k} - 1$ .

[2]

Deduce a similar expression for $v (T + k)$ in terms of $k$ .

[2]

Hence, show that $v (T - k) + v (T + k) \geq 0$ .

[3]

Markscheme

$\frac{d v}{d t} = - (1 + v)$ (A1)

$\int 1 d t = \int - \frac{1}{1 + v} d v$ (or equivalent / use of integrating factor) M1

$t = - \ln (1 + v) (+ C)$ A1

EITHER

attempt to find $C$ with initial conditions $t = 0, v = v_{0}$ M1

$C = \ln (1 + v_{0})$

$t = \ln (1 + v_{0}) - \ln (1 + v)$

$t = \ln (\frac{1 + v_{0}}{1 + v}) \Rightarrow e^{t} = \frac{1 + v_{0}}{1 + v}$ A1

$e^{t} (1 + v) = 1 + v_{0}$

$1 + v = (1 + v_{0}) e^{- t}$ A1

$v (t) = (1 + v_{0}) e^{- t} - 1$ AG

$t - C = - \ln (1 + v) \Rightarrow e^{t - C} = \frac{1}{(1 + v)}$

Attempt to find $C$ with initial conditions $t = 0, v = v_{0}$ M1

$e^{- C} = \frac{1}{(1 + v_{0})} \Rightarrow C = \ln (1 + v_{0})$

$t - \ln (1 + v_{0}) = - \ln (1 + v) \Rightarrow t = \ln (1 + v_{0}) - \ln (1 + v)$

$t = \ln (\frac{1 + v_{0}}{1 + v}) \Rightarrow e^{t} = \frac{1 + v_{0}}{1 + v}$ A1

$e^{t} (1 + v) = 1 + v_{0}$

$1 + v = (1 + v_{0}) e^{- t}$ A1

$v (t) = (1 + v_{0}) e^{- t} - 1$ AG

$t - C = - \ln (1 + v) \Rightarrow e^{- t + C} = 1 + v$ A1

$k e^{- t} - 1 = v$

Attempt to find $k$ with initial conditions $t = 0, v = v_{0}$ M1

$k = 1 + v_{0}$

$e^{- t} (1 + v_{0}) = 1 + v$ A1

$v (t) = (1 + v_{0}) e^{- t} - 1$ AG

Note: condone use of modulus within the ln function(s)

[6 marks]

recognition that when $t = T, v = 0$ M1

$(1 + v_{0}) e^{- T} - 1 = 0 \Rightarrow e^{- T} = \frac{1}{1 + v_{0}}$ A1

$e^{T} = 1 + v_{0}$ AG

Note: Award M1A0 for substituting $v_{0} = e^{T} - 1$ into $v$ and showing that $v = 0$ .

[6 marks]

b.i.

$s (t) = \int v (t) d t (= \int ((1 + v_{0}) e^{- t} - 1) d t)$ (M1)

$= - (1 + v_{0}) e^{- t} - t (+ D)$ A1

( $t = 0, s = 0$ so) $D = 1 + v_{0}$ A1

$s (t) = - (1 + v_{0}) e^{- t} - t + 1 + v_{0}$

at $s_{max}, e^{T} = 1 + v_{0} \Rightarrow T = \ln (1 + v_{0})$

Substituting into $s (t) (= - (1 + v_{0}) e^{- t} - t + 1 + v_{0})$ M1

$s_{max} = - (1 + v_{0}) (\frac{1}{1 + v_{0}}) - \ln (1 + v_{0}) + v_{0} + 1$ A1

$(s_{max} = v_{0} - \ln (1 + v_{0}))$

[5 marks]

b.ii.

METHOD 1

$v (T - k) = (1 + v_{0}) e^{- T} e^{k} - 1$ (M1)

$= (1 + v_{0}) (\frac{1}{1 + v_{0}}) e^{k} - 1$ A1

$= e^{k} - 1$ AG

METHOD 2

$v (T - k) = (1 + v_{0}) e^{- (T - k)} - 1$

$= e^{T} e^{- (T - k)} - 1$ M1

$= e^{T - T + k} - 1$ A1

$= e^{k} - 1$ AG

[2 marks]

METHOD 1

$v (T + k) = (1 + v_{0}) e^{- T} e^{- k} - 1$ (A1)

$= e^{- k} - 1$ A1

METHOD 2

$v (T + k) = (1 + v_{0}) e^{- (T + k)} - 1$ (A1)

$= e^{T} e^{- (T + k)} - 1$

$= e^{T - T - k} - 1$

$= e^{- k} - 1$ A1

[2 marks]

METHOD 1

$v (T - k) + v (T + k) = e^{k} + e^{- k} - 2$ A1

attempt to express as a square M1

$= {(e^{\frac{k}{2}} - e^{- \frac{k}{2}})}^{2} (\geq 0)$ A1

so $v (T - k) + v (T + k) \geq 0$ AG

METHOD 2

$v (T - k) + v (T + k) = e^{k} + e^{- k} - 2$ A1

Attempt to solve $\frac{d}{d k} (e^{k} + e^{- k}) = 0 (\Rightarrow k = 0)$ M1

minimum value of $2$ , (when $k = 0$ ), hence $e^{k} + e^{- k} \geq 2$ R1

so $v (T - k) + v (T + k) \geq 0$ AG

[3 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

Syllabus sections

Topic 5 —Calculus » SL 5.9—Kinematics problems

Show 91 related questions

18M.2.SL.TZ2.S_9a:
Find the initial velocity of P.
18M.2.SL.TZ2.S_9d:
Find the acceleration of P when it changes direction.
19M.1.SL.TZ1.S_7b:
Find the total distance travelled in the first 5 seconds.
19M.2.SL.TZ2.S_8c:
Find the value of $t$ when particle A first changes direction.
22M.2.SL.TZ1.5b:
Find the acceleration of the particle when it changes direction.
21N.1.SL.TZ0.7b:
Sketch a graph of $v$ against $t$ , clearly showing any points of intersection with the axes.
18M.2.SL.TZ1.S_10a:
Find the coordinates of A.
18M.2.SL.TZ1.S_10b.i:
For the graph of $f$ , write down the amplitude.
18M.2.AHL.TZ2.H_7b.ii:
Find the value of the acceleration of P at time t₁.
16N.2.SL.TZ0.S_9b:
Find the value of $p$ .
17M.2.SL.TZ1.S_7a.i:
Write down the first value of $t$ at which P changes direction.
17M.2.SL.TZ1.S_7a.ii:
Find the total distance travelled by P, for $0 \leqslant t \leqslant 8$ .
17M.2.AHL.TZ1.H_11c:
Determine the value of $h$ .
17N.2.SL.TZ0.S_9b:
Hence or otherwise, find all possible values of $t$ for which the velocity of P is decreasing.
18M.2.SL.TZ2.S_9e:
Find the total distance travelled by P.
SPM.2.SL.TZ0.6a:
Find the maximum distance of the particle from O.
19N.2.SL.TZ0.S_10c:
During the second stage, the rocket accelerates at a constant rate. The distance which the rocket travels during the second stage is the same as the distance it travels during the first stage.

Find the total time taken for the two stages.
18M.2.SL.TZ1.S_10e:
Find the first time when the ball’s speed is changing at a rate of 2 cm s⁻².
18M.3.AHL.TZ0.Hca_4a:
Show that $f'\left( 0 \right) = 0$ .
19N.2.AHL.TZ0.H_9b.i:
the value of ${t_1}$ .
19M.2.SL.TZ2.S_8d:
Find the total distance travelled by particle A in the first 3 seconds.
18N.2.SL.TZ0.S_4a:
Find when the particle is at rest.
17M.2.AHL.TZ1.H_11b:
Calculate the vertical distance Xavier travelled in the first 10 seconds.
16N.2.SL.TZ0.S_9a:
Find the initial velocity of $P$ .
18M.2.AHL.TZ2.H_7b.i:
Find an expression for the acceleration of P at time t.
SPM.2.SL.TZ0.6b:
Find the acceleration of the particle at the instant it first changes direction.
18M.2.SL.TZ1.S_10d:
Find the maximum speed of the ball.
18M.3.AHL.TZ0.Hca_4c:
Hence show that the Maclaurin series for $f\left( x \right)$ up to and including the term in ${x^4}$ is ${x^2} + \frac{1}{3}{x^4}$ .
18M.2.SL.TZ2.S_9b:
Find the maximum speed of P.
18M.2.AHL.TZ2.H_7a:
Determine the first time t₁ at which P has zero velocity.
18M.3.AHL.TZ0.Hca_4b:
By differentiating the above equation twice, show that

$\left( {1 - {x^2}} \right){f^{\left( 4 \right)}}\left( x \right) - 5x{f^{\left( 3 \right)}}\left( x \right) - 4f''\left( x \right) = 0$

where ${f^{\left( 3 \right)}}\left( x \right)$ and ${f^{\left( 4 \right)}}\left( x \right)$ denote the 3rd and 4th derivative of $f\left( x \right)$ respectively.
16N.2.SL.TZ0.S_9c:
(i) Find the value of $q$ .

(ii) Hence, find the speed of P when $t = q$ .
19N.2.AHL.TZ0.H_9a.ii:
Hence, write down the maximum speed of the body.
18M.3.AHL.TZ0.Hca_4d:
Use this series approximation for $f\left( x \right)$ with $x = \frac{1}{2}$ to find an approximate value for ${\pi ^2}$ .
17N.1.AHL.TZ0.H_5:
A particle moves in a straight line such that at time $t$ seconds $(t \geqslant 0)$ , its velocity $v$ , in ${\text{m}}{{\text{s}}^{ - 1}}$ , is given by $v = 10t{{\text{e}}^{ - 2t}}$ . Find the exact distance travelled by the particle in the first half-second.
19M.2.SL.TZ2.S_8e.i:
Given that particles A and B start at the same point, find the displacement function ${s_{\text{B}}}$ for particle B.
17M.1.AHL.TZ2.H_4b:
Find the displacement of the particle when $t = {t_1}$
22M.2.SL.TZ1.5c:
Find the total distance travelled by the particle.
19N.2.AHL.TZ0.H_9b.ii:
the distance travelled between $t = 0$ and $t = {t_1}$ .
EXN.2.SL.TZ0.3a:
Find the value of $T$ .
20N.2.AHL.TZ0.H_11e.i:
Show that, at these times, $\tan 6 t = 2$ .
20N.2.AHL.TZ0.H_11c:
Find the maximum displacement of $P$ , in metres, from its initial position.
19N.2.AHL.TZ0.H_9a.i:
Determine the coordinates of point ${\text{A}}$ and the coordinates of point ${\text{B}}$ .
17N.2.SL.TZ0.S_9a:
Write down the values of $t$ when $a = 0$ .
17M.2.AHL.TZ1.H_11a:
Find his velocity when $t = 15$ .
21M.2.SL.TZ1.5c:
Find the acceleration of the particle when $t = 7$ .
21M.1.AHL.TZ2.11d:
Deduce a similar expression for $v (T + k)$ in terms of $k$ .
19M.2.SL.TZ2.S_8a:
Find the initial displacement of particle A from point P.
20N.2.AHL.TZ0.H_11b:
Find an expression for $s$ in terms of $t$ .
20N.1.SL.TZ0.S_7b:
Particle $P_{2}$ also moves in a straight line. The position of $P_{2}$ is given by $r = (\begin{matrix} - 1 \\ 6 \end{matrix}) + t (\begin{matrix} 4 \\ - 3 \end{matrix})$ .

The speed of $P_{1}$ is greater than the speed of $P_{2}$ when $t > q$ .

Find the value of $q$ .
18M.2.SL.TZ1.S_10b.ii:
For the graph of $f$ , write down the period.
19M.2.SL.TZ2.S_8e.ii:
Find the other value of $t$ when particles A and B meet.
17M.1.AHL.TZ2.H_4a:
Find ${t_1}$ and ${t_2}$ .
21M.1.AHL.TZ2.11a:
By solving an appropriate differential equation, show that the particle’s velocity at time $t$ is given by $v (t) = (1 + v_{0}) e^{- t} - 1$ .
21M.1.SL.TZ2.9b.i:
Find the value of $q$ .
18M.2.SL.TZ2.S_9c:
Write down the number of times that the acceleration of P is 0 m s⁻² .
19N.2.AHL.TZ0.H_9b.iii:
the acceleration when $t = {t_1}$ .
21M.1.AHL.TZ2.11c:
By using the result to part (b) (i), show that $v (T - k) = e^{k} - 1$ .
17N.2.SL.TZ0.S_9d:
Find the total distance travelled by P when its velocity is increasing.
18N.2.SL.TZ0.S_4b:
Find the acceleration of the particle when $t = 2$ .
21M.1.AHL.TZ2.11b.ii:
By solving an appropriate differential equation and using the result from part (b) (i), find an expression for $s_{max}$ in terms of $v_{0}$ .
19N.2.SL.TZ0.S_10b:
Find the distance that the rocket travels during the first stage.
19N.2.AHL.TZ0.H_9c:
Find the distance travelled in the first 30 seconds.
EXN.2.SL.TZ0.3b:
Let $s_{1}$ be the distance travelled by the particle from $t = 0$ to $t = T$ and let $s_{2}$ be the distance travelled by the particle from $t = T$ to $t = 1$ .

Show that $s_{2} > s_{1}$ .
21M.1.SL.TZ2.9b.ii:
Find the displacement of particle A from the origin when $t = q$ .
21M.1.SL.TZ2.9d:
Find the value of $d$ .
21N.1.SL.TZ0.7c:
Find the total distance travelled by $P$ .
21N.1.SL.TZ0.7a.ii:
Show that the distance of $P$ from $O$ at this time is $\frac{88}{27}$ metres.
22M.2.SL.TZ1.5a:
Find the value of $t$ when the particle is at rest.
22M.2.SL.TZ2.6a:
Determine when the particle changes its direction of motion.
19M.2.SL.TZ2.S_8b:
Find the value of $t$ when particle A first reaches point P.
17N.2.SL.TZ0.S_9c:
Find an expression for the velocity of P at time $t$ .
18M.2.SL.TZ1.S_10c:
Hence, write $f\left( x \right)$ in the form $p\,\,{\text{cos}}\,\left( {x + r} \right)$ .
19N.2.SL.TZ0.S_10a:
Find an expression for the velocity, $v$ m s⁻¹, of the rocket during the first stage.
17M.2.SL.TZ2.S_7:
Note: In this question, distance is in metres and time is in seconds.

A particle moves along a horizontal line starting at a fixed point A. The velocity $v$ of the particle, at time $t$ , is given by $v(t) = \frac{{2{t^2} - 4t}}{{{t^2} - 2t + 2}}$ , for $0 \leqslant t \leqslant 5$ . The following diagram shows the graph of $v$

There are $t$ -intercepts at $(0,{\text{ }}0)$ and $(2,{\text{ }}0)$ .

Find the maximum distance of the particle from A during the time $0 \leqslant t \leqslant 5$ and justify your answer.
20N.2.AHL.TZ0.H_11a:
Find the times when $P$ comes to instantaneous rest.
20N.2.AHL.TZ0.H_11e.ii:
Hence show that $\frac{v_{2}}{v_{1}} = \frac{v_{3}}{v_{2}} = - e^{- \frac{π}{2}}$ .
21M.2.SL.TZ1.5b:
Find the total distance travelled by the particle.
20N.2.AHL.TZ0.H_11d:
Find the total distance travelled by $P$ in the first $1.5$ seconds of its motion.
20N.1.SL.TZ0.S_7a:
Find an expression for the velocity of $P_{1}$ at time $t$ .
17M.2.SL.TZ1.S_7b:
A second particle Q also moves along a straight line. Its velocity, ${v_{\text{Q}}}{\text{ m}}\,{{\text{s}}^{ - 1}}$ after $t$ seconds is given by ${v_{\text{Q}}} = \sqrt t$ for $0 \leqslant t \leqslant 8$ . After $k$ seconds Q has travelled the same total distance as P.

Find $k$ .
19M.2.AHL.TZ2.H_6:
A particle moves along a horizontal line such that at time $t$ seconds, $t$ ≥ 0, its acceleration $a$ is given by $a$ = 2 $t$ − 1. When $t$ = 6 , its displacement $s$ from a fixed origin O is 18.25 m. When $t$ = 15, its displacement from O is 922.75 m. Find an expression for $s$ in terms of $t$ .
18N.2.SL.TZ0.S_4c:
Find the total distance travelled by the particle.
19M.1.SL.TZ1.S_7a:
Find the value of $s\left( 4 \right) - s\left( 2 \right)$ .
16N.2.SL.TZ0.S_9d:
(i) Find the total distance travelled by P between $t = 1$ and $t = p$ .

(ii) Hence or otherwise, find the displacement of P from A when $t = p$ .
21M.2.SL.TZ1.5a:
Find the smallest value of $t$ for which the particle is at rest.
21M.1.SL.TZ2.9e:
A second particle, particle B, travels along the same straight line such that its velocity is given by $v (t) = 14 - 2 t$ , for $t \geq 0$ .

When $t = k$ , the distance travelled by particle B is equal to $d$ .

Find the value of $k$ .
21M.1.SL.TZ2.9a:
Find the value of $p$ .
21M.1.SL.TZ2.9c:
Find the distance of particle A from the origin when $t = 10$ .
21M.1.AHL.TZ2.11b.i:
Show that the time $T$ taken for the particle to reach $s_{max}$ satisfies the equation $e^{T} = 1 + v_{0}$ .
21N.1.SL.TZ0.7a.i:
Find the value of $t$ when $P$ reaches its maximum velocity.

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Topic 5 —Calculus

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