Question #8fa22

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Question #8fa22

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Answer 1 · 2016-03-04T21:50:40

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

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

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

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

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Question #8fa22

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