# Question #8fa22

Mar 4, 2016

Can I check if the answer for the final part is actually $\left(100 + \frac{2}{5 \mu}\right)$? 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 $\mu \cdot N = \mu \cdot m g$, (N being the normal contact force). Hence by Newton's 2nd law,
$F = m a = \mu \cdot m g$

Hence the maximum acceleration $a = \mu \cdot g = 10 \cdot \mu$ taking $g$ to be $10 m {s}^{-} 2$

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
$v = u + a t$
we get to
$8 = 0 + 10 \mu \cdot t$
so
$t = \frac{8}{10 \mu}$
and distance covered in this time whilst accelerating:
$s = \frac{1}{2} \left(u + v\right) t = \frac{1}{2} \cdot \left(0 + 8\right) \frac{8}{10 \mu} = \frac{32}{10 \mu}$
Hence the time whilst running at constant 8m/s is given by

$t = \frac{s}{v} = \frac{800 - \frac{32}{10 \mu}}{8} = 100 - \frac{4}{10 \mu}$
so (nearly there!) total time is given by:
$100 - \frac{4}{10 \mu} + \frac{8}{10 \mu} = 100 + \frac{2}{5 \mu}$