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AQA GCSE · Question 07.1 · Forces
Figure 12 shows a velocity-time graph for a train travelling between two stations. Determine the distance travelled by the train in the first 600 s of the journey.
Figure 12 shows a velocity-time graph for a train travelling between two stations. Determine the distance travelled by the train in the first 600 s of the journey.
How to approach this question
1. Recall that the distance travelled is the area under a velocity-time graph.
2. Identify the area under the graph between t=0 and t=600 s.
3. This area is a trapezium. You can calculate its area directly using the formula A = ½(a+b)h, or you can split it into a triangle and a rectangle.
4. Read the values for the base(s) and height from the graph and substitute them into the formula(s).
Full Answer
The distance travelled by an object is represented by the area under its velocity-time graph. To find the distance in the first 600 seconds, we need to calculate the area under the graph from t=0 to t=600.
This shape is a trapezium. We can split it into two simpler shapes: a triangle for the acceleration phase and a rectangle for the constant velocity phase.
From the graph:
- Acceleration phase (triangle): from t=0 to t=250 s. The velocity increases from 0 to 58 m/s.
- Constant velocity phase (rectangle): from t=250 s to t=650 s. The velocity is constant at 58 m/s. We are interested up to t=600 s.
1. **Area of the triangle (0 to 250 s):**
Area = ½ × base × height
Area = ½ × 250 s × 58 m/s = 7250 m
2. **Area of the rectangle (250 to 600 s):**
Area = width × height
Width = 600 s - 250 s = 350 s
Area = 350 s × 58 m/s = 20300 m
3. **Total Distance:**
Total distance = 7250 m + 20300 m = 27550 m
Common mistakes
✗ Misreading the scales on the axes (e.g., reading the top speed as 60 m/s instead of 58 m/s, or the time for acceleration as 200s instead of 250s).\n✗ Calculating the gradient instead of the area.\n✗ Using an incorrect formula for the area of a triangle or rectangle.
Practice the full AQA GCSE Physics Higher Tier Paper 2
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