Jane and Dick compete in a 1.00km race. Jane’s strategy is to run the first 600m

Question

Jane and Dick compete in a 1.00km race. Jane’s strategy is to run the first 600m of the race at a constant velocity of 4.00m/s, and then to accelerate to her maximum velocity of 7.50m/s, which takes her 60.0s, and then to finish the race at that velocity. Dick, however, decides to accelerate to his maximum velocity of 8.00m/s at the start of the race and to maintain that speed throughout the rest of the race. Dick takes 180s to reach his maximum velocity. Assuming all accelerations are constant, who wins the race?

Explanation / Answer

Total distance = 1000 m

For jane:
t1 = distance/speed = 600/4 = 150 s
t2 = 60 s
in time t2, distance ran = d2 m
use:
vf=vi+a*t
7.5 = 4 + a*60
a = 0.0583 m/s^2

vf^2 = vi^2 + 2*a*d2
7.5^2 = 4^2 + 2*(0.0583)*d2
d2= 345.2 m

d3 = 1000 - d1 -d2 = 1000 - 600-345.2 = 54.8 m
t3 = distance/speed = 54.8/7.5 = 7.3 s
Total time taken = t1+t2+t3
= 150+60 + 7.3
= 217.2 s

for Dick:
While accelerating:
t1 = 180 s
vf = a*t1
8 = a*180
a = 0.044 m/s^2

d1 = 0.5*a*t1^2
=0.5*(0.044)*(180)^2
= 720 m

remaining distance, d2 = 1000 - d1
= 1000 - 720
= 280 m
t2 = distance/speed = 280/8 = 35 s
Total time taken = t1+t2
= 180 + 35
= 215 s

Dick takes 215 s whereas Jane takes 217.2 s to complete the race
Clearly Dick is the winner

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