Fig. 6.1 shows a velocity-time graph for a bus journey. (a) Calculate the acceleration of the bus during the first 10 s. [2] (b) Calculate the total distanc...
Question 1 Report
Fig. 6.1 shows a velocity-time graph for a bus journey.
(a) Calculate the acceleration of the bus during the first 10 s. [2]
(b) Calculate the total distance travelled by the bus. [2]
(c) Calculate the deceleration of the bus as it comes to a stop. [2]
Answer Details
(a) Acceleration during the first 10 s [2]
From the velocity-time graph, the bus accelerates from 0 to 15 m/s in the first 10 s:
\( a = \frac{\Delta v}{\Delta t} = \frac{15 - 0}{10} \) [1]
\( a = 1.5 \text{ m/s}^2 \) [1]
The straight line from the origin shows uniform (constant) acceleration.
(b) Total distance travelled [2]
The total distance is found from the area under the velocity-time graph. The graph forms a trapezium (or can be split into three regions):
- Region 1 (0 to 10 s): triangle, area = \( \tfrac{1}{2} \times 10 \times 15 = 75 \text{ m} \)
- Region 2 (10 to 30 s): rectangle, area = \( 20 \times 15 = 300 \text{ m} \)
- Region 3 (30 to 40 s): triangle, area = \( \tfrac{1}{2} \times 10 \times 15 = 75 \text{ m} \)
Total distance = 75 + 300 + 75 = 450 m [2]
(c) Deceleration as the bus stops [2]
From the graph, the bus decelerates from 15 m/s to 0 between t = 30 s and t = 40 s:
\( \text{deceleration} = \frac{15 - 0}{40 - 30} = \frac{15}{10} \) [1]
\( = 1.5 \text{ m/s}^2 \) [1]
The deceleration happens to equal the initial acceleration, giving the graph a symmetric shape. Deceleration is simply acceleration in the direction opposing the motion.
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