Careful measurements have been made of Olympic sprinters in the 100 meter race.
ID: 1657422 • Letter: C
Question
Careful measurements have been made of Olympic sprinters in the 100 meter race. A simple but reasonably accurate model is that a sprinter accelerate at 3.6 m/s^-2 for 3.33s,
then runs at constant velocity to the finish line.
(a) What is the race time for a sprinter who follows this model?
(b) A sprinter could run a faster race by accelerating faster at the beginning, thus reaching top speed sooner. If a sprinter’s top speed is the same as in part (a), what acceleration would he need to run the 100 meter race in 9.9s?
(c) By what percent did the sprinter need to increase his acceleration in order to decrease his time by 1%?
Explanation / Answer
a) velocity at the end of 3.33 sconds = 3.6 ( 3.33) = 11.988 m/s
displacement covered in 3.33 second s= 1/2 ( 3.6) ( 3.33)^2= 19.96 m apprx
distance remaining = 100- 19.96= 80.04 m apprx
time to cover remaining distance = 80.04/11.988 = 6.676 seconds apprx
total race time=6.676 + 3.33= 10.006 seconds apprx
b) let the sprinter accelerate for "t" seconds and then run with constant velocity (11.988 m/s) for 9.9 - t
100 = 1/2 a( t^2 ) + 11.988 m/s ( 9.9-t)
a= ( 12/t)
100 = 0.5 ( 12t) - 11.988t + 118.6812
100- 118.6812 =- 5.988 t
t = 3.12 m/s^2
a= 12/ 3.12= 3.846 m/s^2
c) % incraese in accleration = (3.846 - 3.6) / 3.6 x 100= 6.833% apprx