A sprinter in the 100 m dash reaches his top speed of 10 m/s in tau seconds with

Question

A sprinter in the 100 m dash reaches his top speed of 10 m/s in tau seconds with essentially constant acceleration starting from rest. The sprinter then maintains this speed for the remainder of from the race and crosses the finish line 11.5 seconds after starting. [3] sketch the acceleration-time, velocity-time and distance-time curves [1] Determine the average speed of the sprinter for this race. If this sprinter were instead to run in a 200 m race and had the same initial acceleration and top speed, would his average speed for the race increase or decrease?

Explanation / Answer

a]

The slope of the velocity time graph tells us about the acceleration and area under the curve plotted on the velocity time graph tells us about the distance covered during the a certain period of time.

Distance covered during the acceleration period = area under vt curve for t = 0s to t = t seconds.

d = (1/2) x base x height = (1/2)(t)(10) = 5t

So, the remaining distance will be : 100 - 5t

this is covered with a speed of 10 m/s

therefore the time taken to cover this distance will be: (100 - 5t)/10 = 10 - 0.5t

and the total time taken to complete the race is 11.5s

therefore, (10 - 0.5t) + t = 11.5

=> t = 3s

Thus, for a velocity time graph, for time t = 0 to t = 3s, the line has a constant slope which is greater than zero. After this, the velocity remains constant and thus the line on the graph will be horizontal with slope zero.

For acceleration - time graph, the plot will be horizontal for t = 0s to t = 3s and after which the acceleration will be zero.

b] Average speed = Distance/time = 100/11.5 = 8.696 m/s.

c] The acceleration period remains the same as before but the time taken to complete the race will be:

t = 3 + 18.5 = 21.5 s

and so the average speed will be: v = 200/21.5 = 9.3 m/s

which is greater than 8.696 m/s.

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