Further kinematics — WME03 IAL Further Mathematics practice questions · Study-IAL

Further kinematics

WME03Topic 3.13 questions

Motion in a straight line where acceleration depends on time, velocity or displacement.

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Original practice questions written in the style of the exam.

Question 14 marks
A particle $P$ moves along the positive $x$ -axis. When its displacement from $O$ is $x$ metres, its acceleration is $(2x + 3)\,\text{m s}^{-2}$ in the direction of increasing $x$ . When $x = 0$ , $P$ has speed $2\,\text{m s}^{-1}$ .

Find the speed of $P$ when $x = 3$ .
Mark scheme
- M1 Uses $a = v\dfrac{dv}{dx}$ : $\displaystyle\int v\,dv = \int (2x + 3)\,dx$ (1)
Question 25 marks
A particle of mass $2\,\text{kg}$ moves along a smooth horizontal surface. The only horizontal force on it is a resistance of magnitude $4v\,\text{N}$ , where $v\,\text{m s}^{-1}$ is its speed. Initially its speed is .

∫

∫(2x+

3)dx

\tfrac{1}{2}v^2 = x^2 + 3x + c

(1)

M1 Uses

v = 2

x = 0

c = 2

, so

v^2 = 2x^2 + 6x + 4

(1)

v^2 = 40

v = 2\sqrt{10} \approx 6.32\,\text{m s}^{-1}

(1)

10\,\text{m s}^{-1}

(a) Find the time taken for its speed to fall to $5\,\text{m s}^{-1}$ .

(b) Find the distance travelled in that time.

Mark scheme

M1 $2\dfrac{dv}{dt} = -4v$ , separates: $\displaystyle\int \frac{dv}{v} = \int -2\,dt$ (1)
A1 $\ln v = -2t + \ln 10$ , i.e. $v = 10e^{-2t}$ (1)
A1 $5 = 10e^{-2t} \Rightarrow t = \tfrac{1}{2}\ln 2 \approx 0.347\,\text{s}$ (1)
M1 $2v\dfrac{dv}{dx} = -4v \Rightarrow \dfrac{dv}{dx} = -2$ , so (1)
A1 $5 = 10 - 2x \Rightarrow x = 2.5\,\text{m}$ (1)

≈

0.347s

−2

v = 10 - 2x