Edexcel A-Level Mathematics A2 M3.1 Further Kinematics Questions

Practise further kinematics by modelling straight-line motion with variable acceleration, then integrating to connect x, v, t and a.

Exam points

use a = v dv/dx or a = dv/dt to form the differential equation for the motion
integrate v = dx/dt with initial conditions to find t or v in terms of x or t

Covered syllabus points

M3.1.1 - Kinematics of a particle moving

[Maximum number: 7]

A particle P of mass 0.5 kg moves along the positive x-axis in the positive x direction.

At time t seconds, $t \geqslant 1, P$ is x metres from the origin O and is moving with speed $v \mathrm{~m} \mathrm{~s}^{-1}$ . The resultant force acting on P has magnitude $\frac{2}{x^{3}} \mathrm{~N}$ and is directed towards O.

When t=1, x=1 and v=3

Show that

$t=\frac{a+\sqrt{b x^{2}+c}}{d}$ , where a, b, c and d are integers to be found.

$\frac{\mathrm{d} x}{\mathrm{~d} t}=\sqrt{\frac{4}{x^{2}}+5}$ $\int \frac{x}{\sqrt{5 x^{2}+4}} \mathrm{~d} x=\int \mathrm{d} t$ $\frac{1}{5} \sqrt{5 x^{2}+4}=t \quad\left(+k_{1}\right)$ $x=1, t=1 \Rightarrow k_{1}=-\frac{2}{5}$ $t=\frac{2+\sqrt{5 x^{2}+4}}{5}$
(7)
(a)

M1 Dimensionally correct equation of motion. Acceleration must be in form $v \frac{\mathrm{~d} v}{\mathrm{~d} x}$ . Condone missing minus sign.

DM1 Separate variables and attempt integration. Condone missing minus sign.
A1 Correct integrals. Condone missing constant.
DM1 Use v=3, x=1 to find constant of integration. Dependent on both previous M marks.
A1* Reach given result with no errors.

Alt - Definite Integration/Energy Work.
M1 - Equate change in KE to Integral for WD. Integration not needed for this mark and condone inconsistent signs.

\frac{1}{2} \times 2 v^{2}-\frac{1}{2} \times 2 \times 3^{2}=\int( \pm) \frac{2}{x^{3}} d x

DM1 Attempt integration. Condone inconsistent signs.
A1 Correct integration, with consistent signs for KE and WD.
DM1 Substitute in limits.
A1 Reach given result with no errors.
(b)

M1 Use $v=\frac{\mathrm{d} x}{\mathrm{~d} t}$ to form differential equation in x and t.
M1 Separate variables and to produce functions ready to be integrated.
A1 Correct integrands, written in a form that can be integrated.
DM1 Valid attempt to integrate their expression. If they have an incorrect expression, the integration must not be significantly simplified. Dependent on first 2 M marks.

A1 Correct integration. Condone missing constant.
DM1 Use x=1, t=1 to find constant of integration. Dependent on all 3 M marks.
A1 Correct result. (cso)

Question
Number

Scheme

Marks

[Maximum number: 10]

In this question you must show all stages in your working.

Solutions relying entirely on calculator technology are not acceptable.

A particle P is moving along the x-axis.
At time t seconds, where $0 \leqslant t \leqslant \frac{2}{3}, P$ is x metres from the origin O and is moving with velocity $v \mathrm{~m} \mathrm{~s}^{-1}$ in the positive x direction where

v=(2 x+1)^{\frac{3}{2}}

When t=0, P passes through O.

(a)

Find the value of x when the acceleration of P is $243 \mathrm{~m} \mathrm{~s}^{-2}$

[ 4 ]

$a=v \frac{\mathrm{~d} v}{\mathrm{~d} x}$ $=\frac{3}{2}(2 x+1)^{\frac{1}{2}} \times 2 \times(2 x+1)^{\frac{3}{2}}=3(2 x+1)^{2}$ $3(2 x+1)^{2}=243$
x=4
(4)

(b)

Find v in terms of t.

[ 6 ]

$(2x+1)^{3/2}=\frac{\mathrm{d}x}{\mathrm{d}t}$
or $a=3v^{4/3}=\frac{\mathrm{d}v}{\mathrm{d}t}$ $\mathrm{d}t=(2x+1)^{-3/2}\,\mathrm{d}x$ $t=-(2x+1)^{-1/2}+C$ $t=0,\ x=0\Rightarrow C=1$
and obtain an equation in v and t only.
$v=\frac{1}{(1-t)^3}$

[Maximum number: 10]

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

A particle P is moving along a straight line.
At time t seconds, P is a distance x metres from a fixed point O on the line and is moving away from O with speed $\frac{50}{2 x+3} \mathrm{~ms}^{-1}$

(a)

Find the deceleration of P when x=12

Given that x=4 when t=1

[ 5 ]

$\begin{aligned} v &= \frac{50}{2x+3},\\ \frac{dv}{dt} &= \frac{dv}{dx}\frac{dx}{dt}\\ &= \frac{-100}{(2x+3)^2}\times\frac{50}{2x+3}\\ &= \frac{-5000}{(2x+3)^3},\\ x=12:\quad \frac{dv}{dt} &= -\frac{5000}{27^3}=-0.2540\ldots \end{aligned}$
Deceleration $=0.25\ \mathrm{m\,s^{-2}}$ or better.
Notes: use chain rule $\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}$ ;
differentiate with respect to x;
substitute x=12;
deceleration must be positive.

(b)

find the value of t when x=12

[ 5 ]

M1: Use $v=\frac{dx}{dt}$ .
M1: Attempt at integration.
A1: Correct integration, $x^2+3x=50t+c$ , but c may be missing.
A1: Use t=1, x=4 to obtain the correct value of c.
A1: Substitute x=12 to obtain the correct value of t.
ALT 3(b)
Using definite integration: $\int_4^{12}(2x+3)\,dx=\int_1^T 50\,dt$ .
Integrate $[x^2+3x]_4^{12}=[50t]_1^T$ ;
obtain the correct value.

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Covered syllabus points

Question 1

Question 2

Question (a)

Question (b)

Question 3

Question (a)

Question (b)