Question

A 64.9 kg sprinter starts a race with an acceleration of 3.89 m/s2. She keeps this acceleration for 17 m and then maintains the velocity for the remainder of the 100-m dash, what will be her time for the race?

Answer 1

Answer:

Time of race  = 10.18 s

Explanation:

She keeps this acceleration for 17 m and then maintains the velocity for the remainder of the 100-m dash

Time to travel 17 m can be calculated

         s = ut + 0.5at²

         17 = 0 x t + 0.5 x 3.89 x t²

          t = 2.96 s

Velocity after 2.96 seconds

          v = 3.89 x 2.96 = 11.50 m/s

Remaining distance = 100 - 17 = 83 m

Time required to cover 83 m with a speed of 11.50 m/s

          $t=\frac{83}{11.50}=7.22s$

Time of race = 2.96 + 7.22 = 10.18 s