Question

A football player can sprint 40 meters in 4 sec. He has constant acceleration until he reaches his top speed at 10 meters and has constant speed after that. What is his acceleration? What is his constant speed?

Answer 1

1) Acceleration of player is 7.8125
`m/s^(2)`.

2) Constant speed of player is 12.5 m/s.

Step-by-step explanation:

Let the acceleration of the player by 'a' and let he complete the initial 10 meters in
`t_1` time

The initial 10 meters are case of uniformly accelerated motion and hence we can relate the above quantities using second equation of kinematics as

`s=ut+(1)/(2)at^(2)\\\\`

now since the player starts from rest hence u = 0 thus the equation can be written as

`10=(1)/(2)at_1^2\\\\at_(1)^(2)=20...........(i)`

The speed the player reaches after
`t_(1)` time is obtained using first equation of kinematics as

`v=u+at\\v=at_(1)`

Since the total distance traveled by the player is 40 meters hence the total time of trip is 4 seconds hence we infer that he covers 30 meters of distance at a constant speed in time of
`4-t_(1)` seconds

Hence we can write

`(4-t_(1))\cdot at_(1)=30.......(ii)\\\\`

Solving equation i and ii we get

from equation 'i' we obtain
`a=(10)/(t_(1)^(2))`

Using this in equation 'ii' we get

`(4-t_(1))\cdot (20)/(t_(1)^(2))\cdot t_(1)=30\\\\30t_(1)=80-20t_(1)\\\\\therefore t_(1)=(80)/(50)=1.6seconds\\\\\therefore a=(20)/(1.6^2)=7.8125m/s^(2)`

Thus constant speed equals
`v=7.8125* 1.6=12.5m/s`