Linear Kinematics - Physics

Distance and displacement

Some quantities in physics need only a size (a number and a unit). Others need a size and a direction. That single difference splits the whole subject into two families.

Scalar: A quantity with size only, such as distance, speed, mass or time.

Vector: A quantity with size and direction, such as displacement, velocity, acceleration or force.

Distance is how far you have travelled in total, adding up every step of the path. Displacement is how far you end up from your start, measured in a straight line, together with the direction. Walk 3 metres east then 3 metres back west and your distance is \(6\ \text{m}\), but your displacement is \(0\ \text{m}\) because you are back where you began. Both are measured in metres, \( \text{m} \).

The running track trick

A runner who completes one full lap of a 400 m track has travelled a distance of \(400\ \text{m}\) but has a displacement of \(0\ \text{m}\), because the finish line sits exactly on the start line.

Speed and velocity

Speed and velocity are the motion partners of distance and displacement.

Speed: How fast distance is covered. A scalar, measured in metres per second, \( \text{m/s} \).

Velocity: The rate of change of displacement, so it is speed in a stated direction. A vector, also in \( \text{m/s} \).

A car going round a roundabout at a steady \(30\ \text{km/h}\) has a constant speed but a changing velocity, because its direction keeps turning. Whenever direction matters, reach for the vector word.

Average speed

Real journeys speed up and slow down, so we often use an average that smooths the whole trip into one number.

\[ \text{average speed} = \frac{\text{total distance}}{\text{total time}} \]

Worked example

A cyclist covers \(1500\ \text{m}\) in \(120\ \text{s}\). Find the average speed.

Write the values with units: distance \(= 1500\ \text{m}\), time \(= 120\ \text{s}\).

Substitute into \( \text{speed} = \dfrac{\text{distance}}{\text{time}} = \dfrac{1500}{120} \).

Divide: \( 1500 \div 120 = 12.5 \). Carry the unit \( \text{m/s} \).

Average speed \(= 12.5\ \text{m/s}\)

Acceleration

Acceleration measures how quickly velocity changes. If \(u\) is the starting (initial) velocity, \(v\) is the final velocity and \(t\) is the time taken, then

\[ a = \frac{v - u}{t} \]

Acceleration is a vector measured in metres per second squared, \( \text{m/s}^2 \). A positive value means speeding up in the chosen direction; a negative value (sometimes called deceleration) means slowing down.

Worked example

A train speeds up from \(u = 8\ \text{m/s}\) to \(v = 20\ \text{m/s}\) in \(t = 6\ \text{s}\). Find its acceleration.

Find the change in velocity: \( v - u = 20 - 8 = 12\ \text{m/s} \).

Substitute into \( a = \dfrac{v - u}{t} = \dfrac{12}{6} \).

Divide: \( 12 \div 6 = 2 \). The unit is \( \text{m/s}^2 \).

Acceleration \(= 2\ \text{m/s}^2\)

Where this is assessed

Laying out values with units, substituting, then stating the answer with its unit is exactly Criterion C (processing and presenting data). Deciding whether a quantity is a scalar or a vector shows Criterion A (knowing and understanding).

Check yourself

1. A hiker walks \(4\ \text{km}\) north then \(3\ \text{km}\) south. What are the distance and displacement? +

Distance adds up the whole path: \(4 + 3 = 7\ \text{km}\). Displacement is the straight-line result from start to finish: \(4 - 3 = \mathbf{1\ \text{km north}}\), while the distance is \(\mathbf{7\ \text{km}}\).

2. A runner covers \(200\ \text{m}\) in \(25\ \text{s}\). What is the average speed? +

Use \( \text{speed} = \dfrac{\text{distance}}{\text{time}} = \dfrac{200}{25} = \mathbf{8\ \text{m/s}} \).

3. A car slows from \(30\ \text{m/s}\) to \(6\ \text{m/s}\) in \(4\ \text{s}\). Find the acceleration. +

Change in velocity: \( v - u = 6 - 30 = -24\ \text{m/s} \). Then \( a = \dfrac{-24}{4} = \mathbf{-6\ \text{m/s}^2} \). The minus sign shows the car is slowing down.

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