Question

A certain sprinter has a top speed of 11.3 m/s. If the sprinter starts from rest and accelerates at a constant rate, he is able to reach his top speed in a distance of 12.8 m. He is then able to maintain his top speed for the remainder of a 100 m race. (a) What is his time for the 100 m race? (b) In order to improve his time, the sprinter tries to decrease the distance required for him to reach
h his top speed. What must this distance be if he is to achieve a time of 9.75 s for the race?

Answer 1

Answer:

Explanation:

initial velocity u = 0

final velocity v = 11.3 m /s

distance covered s = 12.8 m

v² = u² + 2 a s

11.3² = 0 + 2 x a x 12.8

a = 4.99 m /s²

again ,

v = u + a t

11.3 = 0 + 4.99 t

t = 2.26 s .

Rest of the sprint will be covered with uniform velocity .

Distance covered = 100 - 12.8 = 87.2 m

speed = 11.3 m /s

time taken = 87.2 / 11.3 = 7.7 s

Total time of 100 m sprint = 7.7 + 2.26 = 9.96 m .

b )

Let the time taken to reach the top speed be t .

acceleration a = 11.3 / t

distance covered s = 1/2 a t²

= .5 x (11.3 / t) x t²

= 5.65 t

Rest of the distance = 100 - 5.65 t

time taken to cover rest of the distance = (100 - 5.65 t ) / 11.3

Total time =  (100 - 5.65 t  / 11.3 ) + t = 9.75

100 - 5.65 t + 11.3 t = 11.3 x 9.75

100 + 5.65 t = 110.175

5.65 t = 10.175

t = 1.8

acceleration a = 11.3 / t

= 11.3 / 1.8

= 6.278 m /s²

distance covered in 1.8 s

s = 1/2 a t²

= .5 x 6.278 x 1.8²

= 10.17 m .