Question

How to solve this basic kinematics problem?

A sprinter runs the `100\ m` floors in `10.0\ s`. Approximate its motion assuming a constant acceleration in the first `15\ m` and then a constant speed for the remaining `85\ m`.

How to determinate:
1. its final speed
2. the time taken to cover the first `15\ m`
3. the time needed for the other `85\ m`
4. the magnitude of the acceleration for the first `15\ m`

Answer 1

Final speed
`v=11.5 m/s`
time 1: `t_1 =2.6 s`
time 2:`t_2=7.4 s`
acceleration:
`a=4.41 m/s^2`

Explanation:

You have 4 unknown data (`t_1, t_2,v, a`) and 4 equations:
`t_1 + t_2 = 10.0 s`
`s_2=v xx t_2 =85 m` for the second step
`s_1=1/2 a xx t_1^2= 15 m` for the first step
`v=a xx t_1` for the first step

from the second equation, and by combining the third with the fourth equations, we obtain the time for the 2 paths
`a=v/t_1`
`s_1=1/2 (v/t_1) xx t_1^2= 15 m`
`1/2 v xx t_1= 15 m`
`t_1 = 30/v`
`t_2= 85/v`
that we put in the first equation
`85/v + 30/v =10`

now we have the speed
`85+30 = 10 v`
`v=11.5 m/s`
... the time 1
`t_1 = 30/v= 30/11.5 =2.6 s`
... the time 2
`t_2 = 10-t_1 = 7.4 s`
...the acceleration
`a=v/t_1=(11.5 (m/s/))/(2.6s)=4.41 m/s^2`