Question

A sprinter in a 100m race accelerates from rest at a rate of 5.0m/s^2. She maintains this acceleration for a period of 6.0s. Once she has reached her top speed, she continues to run at her top speed for the rest of the race.

Q. calculate the total time take for the sprinter to run the 100m race.

- how do I get the time while accelerating and time while at constant velocity? Could you please explain how EASY and in detail?

Answer 1

Given Distance (S ) = 100 m

accelaration ( a ) =5.0m/s^-2

She starts running with an accelaration of 5.0m/s^-2 from rest and continues this motion for time , t1= 6 sec

Starting from rest implies at t= 0 initial velocity u = 0  

Step -1

Apply equation of motion to calulate her top ( maximum) speed v during time t1= 6 sec

Note: ( time taken to cover ist part of motion with accelaration is given in question . Use that time t1= 6s to get top velocity )

v= u+ a* t

v=( 0+ 5* 6) m/s. (initial velocity u = 0)

v = 30 m/s

Step 2

Now calculate distance travel during t1= 6sec

using equation of motion

s = u* t+ 1/2 a* t^2 ( where s= distance)

= 0+ 1/2 * 5 * 6^2

s = 90 m

Remaining Distance= (100 - 90) m

= 10 m

Step 3

Calculate time taken to travel remaining Distance = 10 m

with velocity ( top speed) v = 30m/s

Time t = Distance/ speed

Or use this equation of motions (s= u* t + 1/2 a*t^2)

S= u * t ( accelaration becomes zero after t1= 6sec and remaining distance is covered with constant velocity)

t= s/u

t 2= (10 / 30) sec. ( u= 30 m/s that is calculated above)

t2 = 0.3333 sec

Hence total time taken to complete race of 100 m

Total time = t1+ t2

= ( 6+ 0.333) sec

= 6.33s

  

  

Answer 2

SOLUTION :

Distance run  during acceleration period

= ut + 1/2 a t^2

= 0*6 + 1/2 * 5 * 6^2

= 90 m 

Speed attained at the end of acceleration period

= u + a t

= 0 + 5 * 6

= 30 m/s

Distance run at the above constant speed = 100 - 90 = 10 m 

Time taken for this

= distance run at constant speed / Constant speed 

= 10 / 30

= 1/3 = 0.33 sec

So, time taken to run 100m

= time period of acceleration period + time period at constant speed

= 6 + 0.33

= 6.33 sec (ANSWER).