Further Mathematics · Pearson Edexcel 9FM0 · Option FM2 · Topic 5

Further kinematics

Kinematics with variable acceleration, using calculus with displacement, velocity and acceleration.

1 sub-topic
Taught in Year 13
14 practice questions
1 animated lesson

Practice questions (14) Lessons Worksheets All Edexcel Further Mathematics topics

What the specification says

Every sub-topic Pearson Edexcel can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

5.1Further kinematicsY13

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Medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 14 questions in this topic, free, no account needed.

Further kinematics

Not needed for Year 12Medium

Velocity from a variable acceleration · Year 13 · 1 mark

A particle moves in a straight line with acceleration $2t$ m s⁻² at time $t$ seconds. Its velocity is 7 m s⁻¹ when $t = 0$ . What is its velocity when $t = 4$ ?
(1)

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$v$ ${}= \displaystyle\int 2t\,\mathrm{d}t$ ${}= t^2 + c$ , and $c = 7$ . $C$ : 23 m s⁻¹

Total1

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

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Further kinematics

Not needed for Year 12Hard

Further kinematics · Year 13 · 7 marks

A boat moving in a straight line switches off its engine when its speed is 15 m s⁻¹. Its deceleration is then

0.05v^2

m s⁻², where

v

is its speed.

(a)
Show that $v = \dfrac{15}{1 + 0.750t}$ , where $t$ is the time since the engine was switched off.
(4)
(b)
Find the speed of the boat and the distance it has travelled after 2 s.
(3)

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Part (a)4 marks

$\frac{\mathrm{d}v}{\mathrm{d}t}$ ${}= -0.05v^2$
Separate the variables. $\int v^{-2}\,\mathrm{d}v$ ${}= -\int 0.05\,\mathrm{d}t$ ${}\Rightarrow -\frac{1}{v}$ ${}= -0.05t + c$
$t = 0, v = 15$

Part (b)3 marks

$v = 6.00\text{ m s}^{-1}$
$x$ ${}= \int_0^{2} \frac{15}{1 + 0.750t}\,\mathrm{d}t$ ${}= \frac{1}{0.05}\ln(1 + 1.50)$
$= 18.3\text{ m}$

Total7

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

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