A motorcyclist moving with uniform retardation takes 10s and 20s to travel successive quarter kilometer.How much further he will travel before coming to rest?

1 Answer
Feb 22, 2018

Let the motorcyclist be moving with initial velocity #v_0# and with uniform retardation #a# m/s. He takes 10s to travel 250 m and 30s to travel 500m from his starting point.

So #250=v_0*10-a/2*10^2#

#=>25=v_0-a*5......(1)#

Again

#500=v_0*30-a/2*30^2#

#50/3=v_0-a*15....(2)#

Subtracting (2) from (1) we get

#10a=25-50/3=25/3#

#=>a=2.5/3ms^-2#

Inserting #a=2.5/3# in (1) we have

#25=v_0-2.5/3*5#

#=>v_0=25+12.5/3=87.5/3 # m/s

If he travels total #s# m before coming to rest then

#0^2=v_0^2-2*a*s#

#=>0^2=(87.5/3)^2-2*2.5/3*s#

#=>s=87.5^2/9xx3/5=1531.25/3# m

So he travels in last phase before coming to rest

#=1531.25/3-500=31.25/3~~10.42m#