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Motion Graphs

A graph turns a whole journey into a single picture. Once you know that a gradient is a rate and an area is a total, distance-time and velocity-time graphs tell you almost everything about the motion.

MYP 4PhysicsMotionCriteria C · D~10 min read

Written by Fayek · MYP student

Distance-time graphs

On a distance-time graph, time goes along the bottom (x-axis) and distance goes up the side (y-axis). The steepness of the line, its gradient, tells you the speed.

\[ \text{gradient} = \frac{\text{change in distance}}{\text{change in time}} = \text{speed} \]

Reading the shape:

What the line doesWhat the motion is
Flat (horizontal)Stationary, distance is not changing
Straight and slopedSteady (constant) speed
Steeper straight lineFaster steady speed
Curve getting steeperSpeeding up (accelerating)

Velocity-time graphs

Here the y-axis shows velocity instead of distance. The gradient now means something different: it is the acceleration.

\[ \text{gradient} = \frac{\text{change in velocity}}{\text{change in time}} = \text{acceleration} \]
What the line doesWhat the motion is
Flat (horizontal)Constant velocity, zero acceleration
Sloping upAccelerating (speeding up)
Sloping downDecelerating (slowing down)
Line on the time axisAt rest, velocity is zero

Do not mix the two up

A flat line on a distance-time graph means stopped. A flat line on a velocity-time graph means moving at a steady speed. Always check the y-axis label first.

Area under the line

On a velocity-time graph there is a second useful feature: the area under the line equals the distance travelled. For a rectangle that is just velocity times time; for a triangle it is \( \tfrac{1}{2} \times \text{base} \times \text{height} \).

\[ \text{distance} = \text{area under the velocity-time graph} \]

Reading a real graph

Worked example

On a velocity-time graph a car goes in a straight line from \(0\ \text{m/s}\) at \(t = 0\ \text{s}\) up to \(20\ \text{m/s}\) at \(t = 5\ \text{s}\). Find (a) the acceleration and (b) the distance travelled.

1
Acceleration is the gradient: \( \dfrac{\text{change in velocity}}{\text{change in time}} = \dfrac{20 - 0}{5 - 0} = \dfrac{20}{5} \).
2
Divide: \( 20 \div 5 = 4 \), so the acceleration is \( 4\ \text{m/s}^2 \).
3
Distance is the area under the line, a triangle: \( \tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} \times 5 \times 20 \).
4
Work it out: \( \tfrac{1}{2} \times 100 = 50 \), so the distance is \( 50\ \text{m} \).
(a) Acceleration \(= 4\ \text{m/s}^2\); (b) distance \(= 50\ \text{m}\)

Where this is assessed

Pulling numbers off a graph, calculating a gradient, and interpreting what the shape means is Criterion C (processing) and Criterion D (reflecting on the science) at once. Always label the axes and read coordinates carefully.

Check yourself

1. A distance-time graph shows a flat horizontal line. What is the object doing? +

Distance is not changing over time, so the object is stationary (at rest). Its speed is \(0\ \text{m/s}\).

2. On a distance-time graph the line rises from \(0\ \text{m}\) to \(60\ \text{m}\) over \(12\ \text{s}\). What is the speed? +

Speed is the gradient: \( \dfrac{60 - 0}{12 - 0} = \dfrac{60}{12} = \mathbf{5\ \text{m/s}} \).

3. A velocity-time graph shows a steady \(10\ \text{m/s}\) for \(8\ \text{s}\) (a flat line). How far does the object travel? +

The area under the line is a rectangle: \( \text{velocity} \times \text{time} = 10 \times 8 = \mathbf{80\ \text{m}} \). The acceleration is zero because the line is flat.

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