Distance–time and velocity–time graphs

On a distance–time graph, the gradient is the speed.

• A steeper line means a faster speed.
• A flat line means the object is stationary (not moving).
• A line going back down means it is returning to the start.

Example: the graph shows Kim's bike ride. Find her speed for the first part in km/h, and how long she stopped.

1. To A: 6 km in 30 minutes, which is 0.5 hours
2. Speed = distance ÷ time = 6 ÷ 0.5 = 12 km/h
3. A to B is flat, from 30 to 45 minutes: 15 minutes

Her return trip, 6 km in 20 minutes, is steeper, so faster: 6 ÷ (20/60) = 18 km/h.

On a velocity–time graph:

• the gradient is the acceleration (negative for slowing down)
• the area under the graph is the distance travelled
• a flat line means a constant velocity.

Example: use the graph to find the acceleration for the first 10 seconds and the total distance travelled.

1. Acceleration = 20 ÷ 10 = 2 m/s²
2. Area of the first triangle: ½ × 10 × 20 = 100 m
3. Rectangle: 20 × 20 = 400 m
4. Last triangle: ½ × 10 × 20 = 100 m
5. Total distance = 600 m

When the graph is a curve, split the area into trapezia (strips with straight tops) to estimate it.

1. Split the area into strips of equal width.
2. Area of each trapezium = ½ × (left height + right height) × width.
3. Add them up.
4. If the curve bends above the straight tops, the estimate is an underestimate; if it bends below, an overestimate.

Example: estimate the distance travelled in the first 6 seconds, using 3 strips.

1. Strips 2 s wide, with heights 0, 10, 16 and 18
2. ½ × (0 + 10) × 2 = 10; ½ × (10 + 16) × 2 = 26; ½ × (16 + 18) × 2 = 34
3. 10 + 26 + 34 = 70 m
4. The curve bends above the strips' tops, so this is an underestimate

A curve's gradient changes as you move along it. To estimate it at a point, draw a tangent: a straight line that just touches the curve there.

1. Draw the tangent at the point, with a ruler.
2. Pick two points far apart on the tangent.
3. Gradient = change in y ÷ change in x.

Example: estimate the acceleration at 2 seconds.

1. The tangent T touches the curve at A(2, 10).
2. T passes through (0, 2) and (4, 18).
3. Gradient = (18 − 2) ÷ (4 − 0) = 16 ÷ 4 = 4 m/s²