Question

An 87.0 kg sprinter starts a race with an acceleration of 2.28 m/s^2. If the sprinter accelerates at that rate for 30 m, and then maintains that velocity for the remainder of the 100 m dash, what will be his time (in s) for the race?

Answer 1

Concept - Use equation of kinematics for one-dimensional motion to find the solution as shown below

Answer 2

SOLUTION :

Let 100 m run be completed in t sec.

Initial speed = 0 m/s

Run for 30 m at constant acceleration 2.28 m/s^2  and then runs at constant speed that got acquired.

Time for 30 m run :

= sqrt ( 2 * distance / acceleration) 

= sqrt((2 * 30 / 2.28)

= 5.13 sec approx. 

Speed after 5.13 sec = 2.28 * 5.13  = 11.70 m/s

Average speed during first 5.13 sec of 30 m run = (11.70 + 0)/2 = 5.85 m/s 

So, we can consider run of 5.13 sec. at average speed of 5.85 m/s and remaining (t - 5.13) sec at the speed of 11.70 m/s^2 .

So, 

Distance run in t sec = 5.85 * 5.13  + 11.70 * (t - 5.13) 

=> 100 =  30 + 11.70 t - 60 

=> 100 = 11.70 t - 30

=> t = (100 + 30) / 11.70 = 11.11 sec.

=> t = time for the race =   11.11 sec  approx. (ANSWER).