Kinematics (SUVAT and variable acceleration) – A-level Maths exam questions with mark scheme

1CoreCalculator3 marks

A car moves along a straight horizontal road with constant acceleration. It passes point $A$ with speed $4\ \mathrm{m\,s^{-1}}$ and, 6 seconds later, passes point $B$ with speed $22\ \mathrm{m\,s^{-1}}$.

(a) Find the acceleration of the car. [1]

(b) Find the distance $AB$. [2]

(a) $3\ \mathrm{m\,s^{-2}}$B1
(b) complete method, e.g. $s=\frac{(4+22)}{2}\times6$ or $s=4\times6+\frac12\times3\times6^2$M1
(b) 78 mA1
Answer: (a) $3\ \mathrm{m\,s^{-2}}$; (b) 78 m
Examiner tip: List $s,u,v,a,t$ first – choosing the SUVAT equation that avoids your own answer to (a) protects you from follow-through errors.

2CoreCalculator3 marks

The velocity–time graph shows the motion of a cyclist moving along a straight road for 14 seconds.

(a) Find the acceleration of the cyclist during the first 4 seconds. [1]

(b) Find the total distance travelled by the cyclist in the 14 seconds. [2]

(a) $\frac{8}{4}=2\ \mathrm{m\,s^{-2}}$B1
(b) area under the graph, e.g. $\frac12(14+6)\times 8$ or $16+48+16$M1
(b) 80 mA1
Answer: (a) $2\ \mathrm{m\,s^{-2}}$; (b) 80 m
Examiner tip: Distance is the area under a velocity–time graph – use the trapezium formula with both parallel sides (14 and 6).

3StandardCalculator5 marks

A ball is projected vertically upwards with speed $14.7\ \mathrm{m\,s^{-1}}$ from a point 2 m above horizontal ground. The ball is modelled as a particle moving freely under gravity.

(a) Find the greatest height above the ground reached by the ball. [2]

(b) Find the time taken for the ball to reach the ground. [3]

(a) $0^2=14.7^2-2\times9.8\times s$ (giving $s=11.025$)M1
(a) $13$ m or 13.0 m (accept awrt 13.0)A1
(b) $-2=14.7t-4.9t^2$ (or equivalent with consistent signs)M1
(b) solves the 3-term quadratic $4.9t^2-14.7t-2=0$M1
(b) $3.1$ s or $3.13$ sA1
Answer: (a) 13.0 m; (b) 3.1 s
Examiner tip: The ball finishes 2 m below its starting point, so the displacement is $-2$, and remember to add the 2 m in (a).

4StandardCalculator6 marks

A train travels along a straight track between two stations. It starts from rest and accelerates uniformly for $T$ seconds until it reaches a speed of $20\ \mathrm{m\,s^{-1}}$. It then travels at a constant speed of $20\ \mathrm{m\,s^{-1}}$ before decelerating uniformly at $0.5\ \mathrm{m\,s^{-2}}$ until it comes to rest at the second station. The whole journey takes 150 seconds and the distance between the stations is 2400 m.

(a) Sketch a speed–time graph for the journey. [2]

(b) Find the value of $T$. [4]

(a) correct shape: straight line from origin, horizontal line, straight line down to the time axisB1
(a) $20$, $T$ and $150$ marked correctly on the axes (allow the end of constant speed marked at $110$ or $150-40$)B1
(b) deceleration time $=\frac{20}{0.5}=40$ sB1
(b) total area equation with correct structure, e.g. $\frac12\times20T+20(110-T)+\frac12\times40\times20=2400$ or trapezium $\frac12(150+(110-T))\times20=2400$M1
(b) correct equation, e.g. $2600-10T=2400$A1
(b) $T=20$A1
Answer: (a) Trapezium-shaped graph: straight line from $(0,0)$ to $(T,20)$, horizontal to $(110,20)$, straight line down to $(150,0)$; (b) $T=20$
Examiner tip: Work out the deceleration time (40 s) first; the constant-speed time is then $150-T-40$, not $150-T$.

5StandardCalculator5 marks

A particle $P$ moves along a straight line. At time $t$ seconds, the acceleration of $P$ is $(4-2t)\ \mathrm{m\,s^{-2}}$. When $t=0$, $P$ is at the point $O$ and has velocity $3\ \mathrm{m\,s^{-1}}$.

(a) Find the velocity of $P$ when $t=2$. [3]

(b) Find the displacement of $P$ from $O$ when $t=3$. [2]

(a) integrates $a$: $v=4t-t^2\,(+c)$ (powers increase by 1)M1
(a) $v=3+4t-t^2$ (uses $v=3$ at $t=0$)A1
(a) $7\ \mathrm{m\,s^{-1}}$A1
(b) integrates their $v$: $s=3t+2t^2-\frac{t^3}{3}$ and substitutes $t=3$M1
(b) 18 mA1
Answer: (a) $7\ \mathrm{m\,s^{-1}}$; (b) 18 m
Examiner tip: SUVAT does not apply when acceleration depends on $t$ – integrate, and use the initial condition to find the constant.

6HardCalculator7 marks

A particle $P$ moves along a straight line. At time $t$ seconds, $t\ge0$, the velocity of $P$ is $v\ \mathrm{m\,s^{-1}}$, where $$v=3t^2-12t+9$$

(a) Find the values of $t$ when $P$ is instantaneously at rest. [2]

(b) Find the acceleration of $P$ when $t=4$. [2]

(a) sets $v=0$ and solves 3TQ, e.g. $3(t-1)(t-3)=0$M1
(a) $t=1$ and $t=3$A1
(b) differentiates: $a=6t-12$M1
(b) $12\ \mathrm{m\,s^{-2}}$A1
(c) integrates: $s=t^3-6t^2+9t$ and evaluates separately on $[0,1]$ and $[1,3]$M1
(c) displacements $4$ and $-4$ (i.e. $s(1)=4$, $s(3)=0$)A1
(c) total distance 8 mA1
Answer: (a) $t=1,\ 3$; (b) $12\ \mathrm{m\,s^{-2}}$; (c) 8 m
Examiner tip: $P$ changes direction at $t=1$: integrating straight from 0 to 3 gives displacement 0, not the distance 8 m.

7HardCalculator6 marks

A motorcycle passes a point $O$ on a straight road, travelling with constant speed $15\ \mathrm{m\,s^{-1}}$. Two seconds later, a police car starts from rest at $O$ and moves along the road in the same direction with constant acceleration $3\ \mathrm{m\,s^{-2}}$.

Find the distance from $O$ at which the police car overtakes the motorcycle. [6]

forms an equation equating the two displacements with a correct 2 s time offset, e.g. $\tfrac12\times3t^{2}=15(t+2)$ where $t$ is the police car's time (or $\tfrac32(T-2)^{2}=15T$ where $T$ is the motorcycle's time)M1
correct equation, e.g. $1.5t^{2}=15t+30$ or $1.5(T-2)^{2}=15T$A1
solves their 3-term quadratic, e.g. $t^{2}-10t-20=0$ or $T^{2}-14T+4=0$M1
$t=5+3\sqrt5\ (\approx11.7)$ or $T=7+3\sqrt5\ (\approx13.7)$, rejecting the other rootA1
substitutes their positive root into either displacement expression to find the distancedM1
awrt $206$ m (accept $105+45\sqrt5$)A1
Answer: approximately 206 m from $O$
Examiner tip: Measure both times from the same instant: the motorcycle has been moving for $t+2$ seconds when the police car has moved for $t$.

8StretchCalculator10 marks

[In this question $\mathbf{i}$ and $\mathbf{j}$ are horizontal unit vectors due east and due north respectively.]

A particle $P$ moves with constant acceleration $(0.5\mathbf{i}+\mathbf{j})\ \mathrm{m\,s^{-2}}$. At time $t=0$, $P$ has velocity $(2\mathbf{i}-3\mathbf{j})\ \mathrm{m\,s^{-1}}$ and is at the point with position vector $(\mathbf{i}+2\mathbf{j})$ m relative to a fixed origin $O$.

(a) Find the speed of $P$ when $t=4$. [3]

(b) Find the value of $t$ when $P$ is moving due east. [2]

(d) Find the value of $t$ when $P$ is moving in a north-easterly direction. [3]

(a) $\mathbf{v}=(2\mathbf{i}-3\mathbf{j})+4(0.5\mathbf{i}+\mathbf{j})$M1
(a) $\mathbf{v}=4\mathbf{i}+\mathbf{j}$A1
(a) speed $=\sqrt{17}$, awrt $4.12\ \mathrm{m\,s^{-1}}$A1
(b) $\mathbf{j}$-component of velocity $=0$: $-3+t=0$ (and $\mathbf{i}$-component $2+0.5t>0$)M1
(b) $t=3$A1
(c) $\mathbf{r}=(\mathbf{i}+2\mathbf{j})+3(2\mathbf{i}-3\mathbf{j})+\frac12\times 9(0.5\mathbf{i}+\mathbf{j})$M1
(c) $(9.25\mathbf{i}-2.5\mathbf{j})$ mA1
(d) equates the components of velocity: $2+0.5t=-3+t$M1
(d) $t=10$A1
(d) velocity $=7\mathbf{i}+7\mathbf{j}$, both components positive, so moving north-east (not south-west)B1
Answer: (a) $\sqrt{17}\approx4.12\ \mathrm{m\,s^{-1}}$; (b) $t=3$; (c) $(9.25\mathbf{i}-2.5\mathbf{j})$ m; (d) $t=10$
Examiner tip: 'Moving due east' means the $\mathbf{j}$-component of velocity is zero – and don't forget to add the starting position in (c).

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