🧰 You need: pen, pencil, ruler, calculator, graph paper or squared paper. Time: 45 minutes. Course: Combined and Separate.
Information – read this first
- Distance–time graph: the gradient (steepness) is the speed. A steeper line means faster. A flat line means the object is stationary.
- Velocity–time graph: the gradient is the acceleration. A flat line means constant velocity. A line sloping down means decelerating.
- The area under a velocity–time graph is the distance travelled.
- distance = speed × time (s = vt) acceleration = change in velocity ÷ time (a = Δv ÷ t)
- Area of a triangle = ½ × base × height. Area of a rectangle = base × height.
The data
Graph A – a student walking to the bus stop
| Time (s) | 0 | 10 | 20 | 30 | 40 | 50 | 60 |
|---|---|---|---|---|---|---|---|
| Distance (m) | 0 | 20 | 40 | 40 | 40 | 70 | 100 |
Graph B – a car between two sets of traffic lights
| Time (s) | 0 | 5 | 10 | 15 | 20 | 25 | 30 |
|---|---|---|---|---|---|---|---|
| Velocity (m/s) | 0 | 10 | 20 | 20 | 20 | 10 | 0 |
What to do
- (2 min) Write the title Motion Graphs and today’s date. Underline both with a ruler.
- (5 min) Copy the information box into your book.
- (10 min) Plot Graph A: time (s) along the bottom from 0 to 60, distance (m) up the side from 0 to 100. Plot the points with small crosses and join them with straight lines using a ruler. Label the three sections: walking at steady speed, stationary (waiting), walking faster.
- (10 min) Plot Graph B: time (s) from 0 to 30, velocity (m/s) from 0 to 20. Join the points with straight lines. Label: accelerating, constant velocity, decelerating. Lightly shade the area under the line and split it into two triangles and a rectangle.
- (15 min) Write the heading Calculations. Copy the worked example, then do Q1–Q6.
- (3 min) Check your answers and correct in a different colour.
Calculations
Worked example (Graph A, 0–20 s): speed = distance ÷ time = 40 ÷ 20 = 2 m/s
Q1 Graph A: what is the student’s speed between 20 s and 40 s?
Show answer
0 m/s – the line is flat, so the student is stationary.
Q2 Graph A: calculate the student’s speed between 40 s and 60 s.
Show answer
Distance = 100 − 40 = 60 m. Speed = 60 ÷ 20 = 3 m/s
Q3 Graph A: calculate the average speed for the whole journey.
Show answer
100 ÷ 60 = 1.7 m/s (to 2 significant figures)
Q4 Graph B: calculate the car’s acceleration in the first 10 s.
Show answer
a = Δv ÷ t = 20 ÷ 10 = 2 m/s²
Q5 Graph B: calculate the car’s deceleration between 20 s and 30 s.
Show answer
Velocity falls by 20 m/s in 10 s. Deceleration = 20 ÷ 10 = 2 m/s²
Q6 Graph B: use the area under the graph to find the total distance the car travels.
Show answer
Triangle: ½ × 10 × 20 = 100 m. Rectangle: 10 × 20 = 200 m. Triangle: ½ × 10 × 20 = 100 m. Total = 400 m
⭐ Challenge
Calculate the car’s average speed for the whole 30 s. Then explain why the car’s average speed is less than its top speed.
Show answer
400 ÷ 30 = 13 m/s (to 2 significant figures). It is lower than the top speed of 20 m/s because the car spends 20 s of the journey speeding up and slowing down, when it is moving slower than 20 m/s.
