Question

Question #8fa22

Answer 1

Can I check if the answer for the final part is actually `(100+2/(5mu))`? if so, please see below (if not either I am wrong or there is a typo in the question!).

Explanation:

By Newton's 3rd law, the force pushing the athlete forward cannot exceed the friction provided by the track (otherwise they will slip), The friction provided by the track is `mu*N=mu*mg`, (N being the normal contact force). Hence by Newton's 2nd law,
`F=ma=mu*mg`

Hence the maximum acceleration `a=mu*g=10*mu` taking `g` to be `10ms^-2`

For a race of 800m where the maximum speed is 8m/s, we first need to consider the time taken whilst accelerating from the standing start of 0m/s to 8m/s.
Using
`v=u + at`
we get to
`8=0+10mu*t`
so
`t=8/(10mu)`
and distance covered in this time whilst accelerating:
`s=1/2(u+v)t=1/2*(0+8)8/(10mu)=32/(10mu)`
Hence the time whilst running at constant 8m/s is given by

`t=s/v=(800-32/(10mu))/8=100-4/(10mu)`
so (nearly there!) total time is given by:
`100-4/(10mu)+8/(10mu)=100+2/(5mu)`