The diagram shows a velocity-time graph for a cyclist riding a short stage of a charity event. Positive velocity is in the direction of the cyclist's motion...
Question 1 Report
The diagram shows a velocity-time graph for a cyclist riding a short stage of a charity event. Positive velocity is in the direction of the cyclist's motion. The cyclist increases speed, rides steadily, then uses the brakes near a water station.
(a) State the maximum velocity of the cyclist. [1]
(b) Calculate the acceleration during the first 4 s. [2]
(c) Calculate the distance travelled during the 18 s shown on the graph. [3]
(d) Explain why the cyclist has a deceleration from 14 s to 18 s. [2]
(e) Which section of the graph shows constant velocity? [2]
Answer Details
(a) The maximum velocity is 6.0 m/s. [1]
(b) The gradient of a velocity-time graph is acceleration:
\[a=\frac{6.0-0}{4.0}=1.5\text{ m/s}^2\]
[2]
(c) Distance is the area under a velocity-time graph.
\[\text{first triangle}=\frac12\times4\times6=12\text{ m}\]
\[\text{rectangle}=10\times6=60\text{ m}\]
\[\text{final triangle}=\frac12\times4\times6=12\text{ m}\]
\[\text{total distance}=12+60+12=84\text{ m}\]
[3]
(d) Braking, and possibly other resistive forces, acts opposite to the cyclist's motion. This gives a resultant force opposite to the velocity, so the velocity decreases. [2]
(e) From 4 s to 14 s the cyclist has constant velocity. The line is horizontal, so the velocity is unchanged. [2]
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