Kinematics: would you help me?

A runner at the Olympics runs a length #s=100m# in #t=12s#. For the first #s_1=20m# he was evenly accelerating and for the rest he was moving with constant velocity . Find his acceleration and velocity he was going at the finish.

1 Answer
Jane
Feb 18, 2018

Suppose,he went on accelerating for #t s#,

so,we can write, #20 = 1/2 a t^2# (from #s=1/2 at^2#,where, #a# is the value of acceleration)

So, #t= sqrt(40/a)#

Now, after going for #t s# with acceleration,if he achieved a final velocity of #v# then he moved his rest of the distance i.e #(100-20)=80 m# with this velocity,and if that took #t' s# then, #80=v*t'#

Now, #t+t'=12#

So, #sqrt(40/a) +80/v = 12#

Again,if he accelerated from rest to achieve a velocity of #v# after going through a distance of #20m# then , #v^2=0+2a*20=40a# or, #v= sqrt(40a)# (from #v^2 =u^2 +2as# here, #u=0#)

So, we can write,

#sqrt(40/a) + 80/(sqrt(40a)) =12#

Solving this we get, #a = 2.5 ms^-2#

And,velocity with which he was going for the rest of his #80m# of journey was #sqrt(40*2.5) =10 m/s#

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