Describing Motion — Speed, Velocity, Acceleration
Speed and Distance
Speed tells you how fast an object is moving. It is a scalar quantity — it has magnitude but no direction. The equation for speed is:
$$v = \frac{s}{t}$$
where \(v\) is speed in m/s, \(s\) is distance in metres (m), and \(t\) is time in seconds (s). This rearranges to \(s = vt\) and \(t = s/v\).
Typical speeds to know: walking ≈ 1.5 m/s, running ≈ 3 m/s, cycling ≈ 6 m/s, car in town ≈ 13 m/s, car on motorway ≈ 30 m/s, train ≈ 55 m/s, sound in air ≈ 330 m/s.
Velocity and Displacement
Velocity is the vector equivalent of speed: it has both magnitude and direction. Two cars travelling at 30 m/s in opposite directions have the same speed but different velocities.
Displacement is similarly the vector equivalent of distance — it is the net distance from the starting point in a specific direction. An athlete who completes one full lap of a 400 m track has travelled 400 m (distance) but has a displacement of 0 m (they ended where they started).
Acceleration
Acceleration is how quickly an object's velocity changes. The equation is:
$$a = \frac{v - u}{t}$$
where \(a\) is acceleration in m/s\(^2\), \(v\) is final velocity in m/s, \(u\) is initial velocity in m/s, and \(t\) is time in seconds. A negative value of \(a\) means the object is decelerating.
A bus accelerates from 4 m/s to 16 m/s in 6 seconds. Calculate the acceleration.
\(a = \frac{v - u}{t} = \frac{16 - 4}{6} = \frac{12}{6} = 2\) m/s\(^2\)
Now suppose the bus then brakes from 16 m/s to rest in 8 s: \(a = \frac{0 - 16}{8} = -2\) m/s\(^2\) (deceleration of 2 m/s\(^2\)).
For uniform (constant) acceleration, use:
$$v^2 = u^2 + 2as$$
where \(s\) is distance travelled in metres. This equation is essential when time is not given or needed.
A car travelling at 20 m/s brakes uniformly to rest over a distance of 50 m. Calculate the deceleration.
Known: \(u = 20\) m/s, \(v = 0\) m/s, \(s = 50\) m.
\(v^2 = u^2 + 2as\)
\(0 = 400 + 2a \times 50\)
\(100a = -400\)
\(a = -4\) m/s\(^2\) (deceleration of 4 m/s\(^2\))
Time was not given and was not needed — this is exactly when \(v^2 = u^2 + 2as\) is the right tool.
Distance-Time Graphs
A distance-time graph shows how far an object has travelled over time. Key features: a horizontal line means stationary; a straight diagonal line means constant speed; a steepening curve means increasing speed (acceleration).
The speed at any point equals the gradient of the line: \(\text{speed} = \frac{\text{change in distance}}{\text{change in time}}\). For a straight line, pick two points far apart and divide the rise by the run. For a curve, draw a tangent at the point of interest and find its gradient.
Velocity-Time Graphs
A velocity-time graph is even more powerful. The gradient gives the acceleration. The area under the line gives the total distance travelled.
A horizontal line means constant velocity (acceleration = 0). A line sloping upward means increasing velocity (positive acceleration). A line sloping downward means decreasing velocity (deceleration). Complex shapes can be split into triangles and rectangles to find the total area (= total distance).
A common method for measuring acceleration is to use a trolley on a ramp with a light gate connected to a data logger. The independent variable is the force applied (e.g. by varying the hanging mass); the dependent variable is the acceleration. Controlled variables include the total mass of the system. The light gate measures the time taken for a card of known width to pass — from this, both speed and acceleration can be calculated. Results plotted as acceleration against force produce a straight line through the origin if \(F = ma\) holds.
Graphs and Calculations: Exam Technique
When interpreting motion graphs, look at the gradient first. A flat line on a distance-time graph means the object is stationary; a flat line on a velocity-time graph means constant velocity with zero acceleration. Steeper gradients indicate faster speed or greater acceleration. To find displacement from a velocity-time graph, calculate the area under the line. At Higher tier this includes reading enclosed areas between the graph line and the time axis, which represent displacement in a particular direction. A common exam error is confusing the gradient, which gives acceleration, with the area, which gives displacement, on a velocity-time graph.