Displacement, velocity and acceleration

AS · 9 min

Kinematics describes motion without asking what causes it. Everything rests on five quantities: distance, displacement, speed, velocity and acceleration. They sound familiar from IGCSE, but at AS the distinction between the scalar and vector versions is examined precisely, and the definitions must be stated in the right words. Get these exact now and the motion graphs, equations of motion and projectiles that follow become straightforward.

Distance and displacement

Imagine walking $300\ \text{m}$ east and then $400\ \text{m}$ north. Your legs have covered $700\ \text{m}$ , but you are only $500\ \text{m}$ from where you started. Those are two different quantities.

Definition

Distance is the total length of path travelled. It is a scalar.

Displacement is the distance moved in a specified direction from a fixed reference point. It is a vector.

Displacement depends only on where you start and where you end, not on the route. A runner who completes one lap of a $400\ \text{m}$ track has run a distance of $400\ \text{m}$ and has a displacement of zero.

In one dimension (a straight line), the direction of a displacement is shown by its sign. Choose a positive direction at the start of every problem and stick to it: if up is positive, a displacement of $-3.0\ \text{m}$ means $3.0\ \text{m}$ below the starting point.

Speed and velocity

Definition

Speed is the rate of change of distance (distance travelled per unit time). It is a scalar.

Velocity is the rate of change of displacement (change in displacement per unit time). It is a vector.

Key result

\text{average speed} = \frac{\text{total distance travelled}}{\text{total time taken}}, \qquad \text{average velocity} = \frac{\text{total displacement}}{\text{total time taken}}

Unit: $\text{m s}^{-1}$ .

Instantaneous speed or velocity is the value at a particular moment: the rate of change over a vanishingly short time interval. On a displacement–time graph it is the gradient at that instant (see Motion graphs). A car's speedometer shows instantaneous speed; average speed for a journey is usually lower.

Uniform (constant) velocity means both the speed and the direction are constant. An object moving in a circle at constant speed does not have constant velocity, because its direction keeps changing.

Distance, displacement, speed and velocity

A student walks $300\ \text{m}$ due east in $4.0$ minutes, then $400\ \text{m}$ due north in $6.0$ minutes. Calculate (a) the distance travelled, (b) the displacement, (c) the average speed and (d) the average velocity.

Solution

(a) Distance $= 300 + 400 = 700\ \text{m}$ .

(b) The two legs are perpendicular, so

s = \sqrt{300^2 + 400^2} = 500\ \text{m}, \qquad \tan\theta = \frac{400}{300} \Rightarrow \theta = 53^\circ \text{ north of east}

\text{average speed} = \frac{700}{600} = 1.17\ \text{m s}^{-1}

(d)

\text{average velocity} = \frac{500}{600} = 0.833\ \text{m s}^{-1} \text{ at } 53^\circ \text{ north of east}

The velocity needs its direction; the speed does not.

Tip

Convert km h $^{-1}$ to m s $^{-1}$ by dividing by $3.6$ : $1\ \text{km h}^{-1} = \dfrac{1000\ \text{m}}{3600\ \text{s}}$ . So $72\ \text{km h}^{-1} = 20\ \text{m s}^{-1}$ and $108\ \text{km h}^{-1} = 30\ \text{m s}^{-1}$ .

Acceleration

Definition

Acceleration is the rate of change of velocity.

Key result

a = \frac{\Delta v}{\Delta t} = \frac{v - u}{t}

where $u$ is the initial velocity and $v$ the final velocity after time $t$ . Unit: $\text{m s}^{-2}$ . Acceleration is a vector.

Because acceleration is the rate of change of velocity, an object accelerates whenever its speed changes or its direction changes. A satellite in a circular orbit moves at constant speed and is accelerating all the time, towards the centre of the circle.

Signs matter. If the positive direction is the direction of motion, a negative acceleration means the object is slowing down (decelerating). But a negative acceleration does not always mean slowing down: a ball falling downwards, with up taken as positive, has $a = -9.81\ \text{m s}^{-2}$ and is speeding up. What matters is whether acceleration and velocity point the same way (speeding up) or opposite ways (slowing down).

Watch out

"Deceleration of $6\ \text{m s}^{-2}$ " and "acceleration of $-6\ \text{m s}^{-2}$ " mean the same thing if the motion is in the positive direction. Do not write "deceleration of $-6\ \text{m s}^{-2}$ ": that is a double negative and means speeding up.

Speeding up and braking

(a) A car accelerates uniformly from $12\ \text{m s}^{-1}$ to $30\ \text{m s}^{-1}$ in $6.0\ \text{s}$ . Find its acceleration. (b) It then brakes uniformly from $25\ \text{m s}^{-1}$ to rest in $4.0\ \text{s}$ . Find the acceleration.

Solution

(a)

a = \frac{v - u}{t} = \frac{30 - 12}{6.0} = 3.0\ \text{m s}^{-2}

(b)

a = \frac{0 - 25}{4.0} = -6.25\ \text{m s}^{-2} \approx -6.3\ \text{m s}^{-2}

The negative sign shows the acceleration is opposite to the velocity: a deceleration of $6.3\ \text{m s}^{-2}$ .

A bouncing ball

A ball falls vertically and hits the floor at $6.0\ \text{m s}^{-1}$ . It rebounds vertically at $4.0\ \text{m s}^{-1}$ . It is in contact with the floor for $0.020\ \text{s}$ . Find the average acceleration during the contact.

Solution

Take upwards as positive. Then $u = -6.0\ \text{m s}^{-1}$ (moving down) and $v = +4.0\ \text{m s}^{-1}$ (moving up).

a = \frac{v - u}{t} = \frac{4.0 - (-6.0)}{0.020} = \frac{10.0}{0.020} = 500\ \text{m s}^{-2} \text{ upwards}

The change in velocity is $10.0\ \text{m s}^{-1}$ , not $2.0\ \text{m s}^{-1}$ . Forgetting that the velocity reverses is the classic error in this question.

Constant speed but changing velocity

Half a circle

A cyclist rides at a constant speed of $10\ \text{m s}^{-1}$ around half of a circular track of radius $50\ \text{m}$ , from point P to the diametrically opposite point Q. Calculate (a) the time taken, (b) the displacement from P to Q, (c) the average velocity, (d) the magnitude of the change in velocity and (e) the magnitude of the average acceleration.

Solution

(a) Distance $= \pi r = \pi \times 50 = 157\ \text{m}$ , so $t = 157/10 = 15.7\ \text{s}$ .

(b) Displacement $=$ the diameter $= 100\ \text{m}$ , in the direction from P to Q.

(c) Average velocity $= 100/15.7 = 6.37\ \text{m s}^{-1}$ from P to Q. It is smaller than the speed because the path is not straight.

(d) At P the velocity is $10\ \text{m s}^{-1}$ in one direction; at Q it is $10\ \text{m s}^{-1}$ in the opposite direction. Change in velocity $= 10 - (-10) = 20\ \text{m s}^{-1}$ .

(e) Average acceleration $= 20/15.7 = 1.27\ \text{m s}^{-2}$ .

The speed never changed, yet the cyclist accelerated throughout.

Summary of the five quantities

Quantity	Scalar or vector	Definition	Unit
distance	scalar	total length of path travelled	$\text{m}$
displacement	vector	distance in a specified direction from a fixed point	$\text{m}$
speed	scalar	rate of change of distance	$\text{m s}^{-1}$
velocity	vector	rate of change of displacement	$\text{m s}^{-1}$
acceleration	vector	rate of change of velocity	$\text{m s}^{-2}$

Exam tip

Definitions are worth a mark each and are marked strictly. "Velocity is speed in a given direction" is usually accepted, but "rate of change of displacement" is the safest. "Acceleration is the change in velocity" (without "per unit time" or "rate of") scores zero.
State your sign convention ("taking upwards as positive") at the start of any calculation involving a reversal of direction. It earns credit for clear working and stops sign errors.
For a vector answer, give a direction, or a sign with the convention stated.
"Explain why an object moving at constant speed can be accelerating" needs: velocity is a vector, its direction changes, so velocity changes, so there is an acceleration.

Summary

Distance and speed are scalars; displacement, velocity and acceleration are vectors.
Displacement is distance in a specified direction from a fixed point; it depends only on start and end positions.
Velocity is the rate of change of displacement; acceleration is the rate of change of velocity.
$a = (v - u)/t$ ; the sign of $a$ relative to $v$ tells you whether the object speeds up or slows down.
A change of direction at constant speed is still an acceleration.
Always choose and state a positive direction before using signs.

Practice

Question

State the difference between speed and velocity.
An athlete runs one and a half laps of a circular track of circumference $400\ \text{m}$ in $75\ \text{s}$ . Calculate the average speed, and the magnitude of the average velocity (the track has a diameter of $127\ \text{m}$ ).
Convert $54\ \text{km h}^{-1}$ to $\text{m s}^{-1}$ , and $15\ \text{m s}^{-1}$ to $\text{km h}^{-1}$ .
A train slows uniformly from $36\ \text{m s}^{-1}$ to $12\ \text{m s}^{-1}$ in $40\ \text{s}$ . Calculate its acceleration.
A tennis ball travelling horizontally at $25\ \text{m s}^{-1}$ is hit straight back at $35\ \text{m s}^{-1}$ . The racket is in contact for $5.0\ \text{ms}$ . Calculate the magnitude of the average acceleration.
A ball is thrown vertically upwards. Taking upwards as positive, state the sign of its velocity and its acceleration (a) on the way up, (b) at the top, (c) on the way down.
Explain why a car going round a roundabout at a steady $8\ \text{m s}^{-1}$ is accelerating.
A boat sails $6.0\ \text{km}$ due north in $30$ minutes, then $8.0\ \text{km}$ due west in $50$ minutes. Calculate its average speed and its average velocity in $\text{km h}^{-1}$ .
A particle moves anticlockwise round a circle of radius $2.0\ \text{m}$ at a constant speed of $3.0\ \text{m s}^{-1}$ . It moves a quarter of a revolution. Calculate the magnitude of (a) its displacement, (b) its change in velocity and (c) its average acceleration over this quarter revolution.
A ball is dropped onto a hard floor and rebounds. Its speed just before impact is $7.0\ \text{m s}^{-1}$ and the magnitude of its average acceleration during the $0.012\ \text{s}$ contact is $1.0 \times 10^{3}\ \text{m s}^{-2}$ . Calculate the rebound speed, and state the direction of the acceleration during contact.

Answers

Speed is the rate of change of distance and is a scalar; velocity is the rate of change of displacement and is a vector (it has a direction).
Distance $= 600\ \text{m}$ , average speed $= 600/75 = 8.0\ \text{m s}^{-1}$ . After one and a half laps the athlete is at the opposite side of the circle, so the displacement is one diameter, $127\ \text{m}$ ; average velocity $= 127/75 = 1.7\ \text{m s}^{-1}$ .
$54/3.6 = 15\ \text{m s}^{-1}$ ; $15 \times 3.6 = 54\ \text{km h}^{-1}$ .
$a = (12 - 36)/40 = -0.60\ \text{m s}^{-2}$ (a deceleration of $0.60\ \text{m s}^{-2}$ ).
Taking the final direction as positive: $u = -25$ , $v = +35$ . $a = (35 - (-25))/(5.0 \times 10^{-3}) = 60/0.0050 = 1.2 \times 10^{4}\ \text{m s}^{-2}$ .
(a) Velocity positive, acceleration negative. (b) Velocity zero, acceleration negative ( $-9.81\ \text{m s}^{-2}$ ; it is not zero at the top). (c) Velocity negative, acceleration negative.
Velocity is a vector. The direction of motion changes continuously, so the velocity changes even though the speed is constant. A changing velocity means an acceleration.
Distance $= 14.0\ \text{km}$ in $80\ \text{min} = 1.333\ \text{h}$ ; average speed $= 10.5\ \text{km h}^{-1}$ . Displacement $= \sqrt{6.0^2 + 8.0^2} = 10.0\ \text{km}$ at $\tan^{-1}(8.0/6.0) = 53^\circ$ west of north; average velocity $= 10.0/1.333 = 7.5\ \text{km h}^{-1}$ at $53^\circ$ west of north.
(a) The start and end points are radii at right angles: $s = \sqrt{2.0^2 + 2.0^2} = 2.83\ \text{m}$ . (b) The velocities are $3.0\ \text{m s}^{-1}$ at right angles: $|\Delta v| = \sqrt{3.0^2 + 3.0^2} = 4.24\ \text{m s}^{-1}$ . (c) Time $= \dfrac{\tfrac{1}{4} \times 2\pi \times 2.0}{3.0} = 1.05\ \text{s}$ , so average acceleration $= 4.24/1.05 = 4.05\ \text{m s}^{-2}$ .
$\Delta v = a t = 1.0 \times 10^{3} \times 0.012 = 12\ \text{m s}^{-1}$ . Taking up as positive, $v - (-7.0) = 12$ , so $v = 5.0\ \text{m s}^{-1}$ upwards. The acceleration during contact is upwards.

Distance and displacement

Speed and velocity

Acceleration

Constant speed but changing velocity

Summary of the five quantities

Summary

Practice

Console