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
Distance is the area under the velocity-time graph.
The shape can be split into a triangle (0-250 s) and a rectangle (250-600 s).
Area of triangle = ½ × base × height = ½ × 250 s × 58 m/s = 7250 m.
Area of rectangle = width × height = (600 - 250) s × 58 m/s = 350 s × 58 m/s = 20300 m.
Total distance = 7250 m + 20300 m = 27550 m.
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.
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