A-level · Kinematics

Motion in two dimensions with vectorsExam-style questions

Why the a → v → r calculus chain works component-by-component in 2D, that speed is the magnitude of the velocity vector, and how "parallel to" and "perpendicular to" a direction become simple component conditions.

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33 questions · A-level level · mark scheme and worked solution on each · no sign-up to try

DifficultyAll Level 2 Level 3 Level 4 Level 5

1
[In this question $\mathbf{i}$ and $\mathbf{j}$ are horizontal unit vectors directed due east and due north respectively and position vectors are given relative to a fixed origin $O$ .]

A coastguard boat $C$ and a yacht $Y$ are each moving with constant velocity, and the boats are modelled as particles. The velocity of $C$ is $(-18\mathbf{i} + 24\mathbf{j})\ \text{km h}^{-1}$ . The yacht $Y$ is moving at $20\ \text{km h}^{-1}$ in the direction of the vector $(3\mathbf{i} - 4\mathbf{j})$ .
(a) Find the velocity of $Y$ , giving your answer in terms of $\mathbf{i}$ and $\mathbf{j}$ .

(2 marks)

At 08:00, $C$ is at the point with position vector $(4\mathbf{i} - 3\mathbf{j})$ km.

At 08:30, $Y$ is at the point with position vector $(-5\mathbf{i} + 11\mathbf{j})$ km.
(b) Find the position vector of $Y$ at time $t$ hours after 08:00, giving your answer in terms of $t$ , $\mathbf{i}$ and $\mathbf{j}$ .

(2 marks)
(c) Show that the position vector of $Y$ relative to $C$ at time $t$ hours after 08:00 is

$\overrightarrow{CY} = [(30t - 15)\mathbf{i} + (22 - 40t)\mathbf{j}]\ \text{km}$

(2 marks)
(d) Show that $C$ and $Y$ do not collide.

(3 marks)
(e) Find the distance between $C$ and $Y$ at the instant when $Y$ is south-east of $C$ .

(4 marks)
Level 5 · 13 marksTry it
2
[In this question $\mathbf{i}$ and $\mathbf{j}$ are horizontal unit vectors directed due east and due north respectively.]

Two particles, $P$ and $Q$ , are moving in a horizontal plane. Each particle has constant acceleration. Time $t$ is measured in seconds.

The acceleration of $P$ is $(2\mathbf{i} - \mathbf{j})\ \text{m s}^{-2}$ . When $t = 0$ , the velocity of $P$ is $(-\mathbf{i} - 2\mathbf{j})\ \text{m s}^{-1}$
(a) Find the speed of $P$ when $t = 2$

(2 marks)

The velocity of $Q$ is $(3\mathbf{i} - \mathbf{j})\ \text{m s}^{-1}$ when $t = 2$ , and $(-\mathbf{i} - 5\mathbf{j})\ \text{m s}^{-1}$ when $t = 6$
(b) Find the acceleration of $Q$ .

(2 marks)
(c) Find the value of $t$ at which $Q$ is moving due east.

(2 marks)
(d) Find the exact value of $t$ , where $t > 0$ , at which $P$ and $Q$ are moving with the same speed.

(4 marks)
(e) Find the value of $t$ at which $P$ and $Q$ are moving in the same direction.

(3 marks)
Level 5 · 13 marksTry it
3
A particle starts at the point with position vector $(\mathbf{i} + 2\mathbf{j})$ m, with velocity $(3\mathbf{i} - \mathbf{j})\ \text{m s}^{-1}$ , and moves with constant acceleration $(2\mathbf{i} + 4\mathbf{j})\ \text{m s}^{-2}$ . Find
(a) its velocity after $2$ s,
(b) its speed after $2$ s,
(c) its position vector after $2$ s, and (d) its distance from the origin $O$ after $2$ s.
Level 5 · 7 marksTry it
4
A particle of mass $0.5$ kg moves with velocity $\mathbf{v} = (t^{2} + 1)\,\mathbf{i} + (3t - 2)\,\mathbf{j}$ m s $^{-1}$ at time $t$ seconds.

Find the magnitude of the resultant force on the particle when $t = 2$ . (5 marks)
Level 5 · 5 marksTry it
5
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

[In this question, $\mathbf{i}$ is a unit vector due east and $\mathbf{j}$ is a unit vector due north.]

A lifeboat is manoeuvring in a bay. It is modelled as a particle $L$ whose position vector relative to a fixed buoy $O$ is $\mathbf{r}$ metres at time $t$ seconds, where

$\mathbf{r} = \left(6t - 3t^{\frac{3}{2}}\right)\mathbf{i} + \left(c\sqrt{t} - 5t\right)\mathbf{j} \qquad 1 \leqslant t \leqslant 12$

and $c$ is a constant.

When $t = 9$ , the lifeboat is on a bearing of $225^\circ$ from the buoy.
(a) Show that $c = 6$

(3 marks)
(b) Find the speed of the lifeboat when $t = 9$

(4 marks)

When $t = T$ , the acceleration of $L$ acts in the direction of the vector $(-9\mathbf{i} - 2\mathbf{j})$ .
(c) Find the value of $T$ .

(4 marks)
Level 5 · 11 marksTry it