15.2 The language of kinematics — Learn

Prerequisites: positive and negative numbers on a number line; using $\text{speed} = \dfrac{\text{distance}}{\text{time}}$; the SI units of length and time from Section 15.1.

By the end of this section you can: describe the position and the displacement of a moving object; tell distance travelled apart from displacement; tell speed apart from velocity; find an average speed, an average velocity and an acceleration; use a sign convention correctly.

1. Position and displacement

Every moving object needs a starting point. Choose a fixed point, called the origin, and a positive direction along a straight line. The position of the object is its place on that line, measured from the origin.

The displacement is the change in position. Displacement is a vector: it has a size and a direction. Along a line, the direction is carried by the sign. Moving in the positive direction gives a positive displacement; moving the other way gives a negative displacement.

If an object starts at position $s_0$ and finishes at position $s_1$, then

Example 1. A cyclist starts at a point $O$ and rides $5$ km east, then $2$ km west. Taking east as positive, find the final position and the displacement.

Start at $0$. Riding east adds $5$ km, and riding west subtracts $2$ km:

The final position is $3$ km east of $O$. The displacement is $3$ km east, or $+3$ km.

2. Distance travelled

The distance travelled is the total length of the path, whatever the direction. It is a scalar, so it has a size but no direction. Distance travelled is never negative.

In Example 1 the cyclist rides $5$ km and then $2$ km, so the distance travelled is

The distance is $7$ km, while the displacement is $3$ km east. They are equal only when the object moves in one direction without turning back.

Example 2. A ball is thrown vertically upwards and caught at the same height. Find the distance travelled and the displacement when the ball has risen $4$ m.

The ball rises $4$ m and falls $4$ m, so the distance travelled is $8$ m. It finishes at the height it started from, so the displacement is $0$ m.

3. Speed and velocity

Speed is the rate at which distance is covered. Velocity is the rate of change of displacement. Velocity is a vector, so it can be positive or negative. The two differ whenever the object turns around.

For motion over a time $t$:

The unit of both is m/s. Speed is the magnitude of velocity, so speed is never negative.

Example 3. A car travels $240$ m due east in $12$ s, then $60$ m due west in $4$ s. Find the average speed and the average velocity over the whole journey.

The distance travelled is $240 + 60 = 300$ m, over a total time of $16$ s:

The displacement is $240 - 60 = 180$ m due east, so

The average speed is larger, because the car doubled back.

4. Acceleration

Acceleration is the rate of change of velocity. Like velocity, it is a vector, so its sign shows its direction along the line.

For an object whose velocity changes from $u$ to $v$ in a time $t$:

The unit is m/s$^2$. A positive acceleration means the velocity is increasing in the positive direction. A negative acceleration means the velocity is changing in the negative direction.

Example 4. A train slows from $30$ m/s to $10$ m/s in $5$ s while travelling east. Taking east as positive, find its acceleration.

The acceleration is $-4$ m/s$^2$. The negative sign means the acceleration points west, opposite to the direction of travel. The train is slowing down.

Example 5. A ball is thrown vertically upwards at $12$ m/s. It rises, stops for an instant and falls back. Taking upwards as positive, write down its acceleration while it is in the air.

Gravity acts downwards throughout, even at the highest point. Taking upwards as positive, the acceleration is $-9.8$ m/s$^2$ at every moment of the flight.

5. Try it: distance and displacement

Predict first. A walker starts at $0$ and walks to a turning point, then walks back to a finishing point. Before you move the sliders, predict when the distance travelled and the displacement are equal. Write down your prediction.

Test. Move the two sliders so the turning point is to the right of the finishing point, then so it is to the left. Watch both readouts.

Explain. Write two sentences. The first says why the displacement can be smaller than the distance travelled. The second says what the readouts show when the walker finishes back at $0$.

Watch out. Three mistakes are common with this vocabulary.

Distance used as displacement. The distance travelled is the length of the path, and the displacement is the change in position. A runner who runs one lap of a track travels $400$ m but has a displacement of $0$ m, because the finish is the starting point.

A negative velocity read as a negative speed. Speed is never negative. A velocity of $-8$ m/s means the object moves at $8$ m/s in the negative direction. The size is $8$ m/s.

A negative acceleration called slowing down. The sign of the acceleration says which way the velocity is changing, not whether the object is slowing. A ball thrown upwards has a negative acceleration on the way up and on the way down. Re-check: decide the positive direction first, then give every vector its sign before you calculate.

You should now be able to

• define position, displacement and distance travelled;
• find a displacement from a starting position and a finishing position;
• find an average speed and an average velocity;
• tell when the average speed and the average velocity of a journey differ;
• find an average acceleration from a change in velocity;
• use a sign convention and interpret a negative velocity or acceleration.