Question #8fa22

1 Answer
Matt
Mar 4, 2016

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)#

Related questions
See all questions in Acceleration
Impact of this question
1347 views around the world
You can reuse this answer
Creative Commons License