The diagram shows two sprinters running in adjacent lanes during a 100 m indoor race. Timing sensors record the time when each runner's torso crosses the fi...
Question 1 Report
The diagram shows two sprinters running in adjacent lanes during a 100 m indoor race. Timing sensors record the time when each runner's torso crosses the finish line. Runner A finishes in 12.4 s. Runner B finishes in 12.0 s, but starts 0.20 s after the starting signal because of a slow reaction. A coach compares average speed with instantaneous speed near the finish.
(a) State which runner has the greater average speed for the 100 m race. [1]
(b) Calculate the average speed of Runner A. [2]
(c) Calculate the time taken by Runner B from starting to crossing the finish line. [3]
(d) Explain why average speed does not show a sprinter's instantaneous speed at the finish line. [3]
(e) Describe one use of a light gate system that could measure instantaneous speed near the finish. [3]
Answer Details
(a) Runner B has the greater average speed because the same 100 m distance is completed in the shorter recorded time. [1]
(b) \[\text{average speed}=\frac{\text{distance}}{\text{time}}=\frac{100}{12.4}=8.06\text{ m/s}\]
The average speed is 8.06 m/s, or 8.1 m/s. [2]
(c) Runner B's time from the signal to the finish is 12.0 s. [1] This includes a reaction delay of 0.20 s. [1]
\[\text{running time}=12.0-0.20=11.8\text{ s}\]
The time spent running is 11.8 s. [3]
(d) Average speed is calculated using the whole 100 m distance [1] and the whole race time. [1] A sprinter's speed changes during a race, so this average may differ from their speed at the single instant they cross the finish line. [1]
(e) Place two light gates a known short distance apart near the finish. [1] Measure the time taken for the runner's torso or a card to pass from one gate to the other. [1] Calculate \(\text{speed}=\text{distance between gates}/\text{short time}\). [1]
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