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

^{2} " , where u is initial velocity, a is retardation and t is time duration.

for first quarter kilometer travel in 10 s:- 250 = u×10 - (1/2)×a×100 or u - 5 a = 25 …(1)

first and second qurter kilometer travel in 30 s: - 500 = u×30 -(1/2)×a×900 or 3u - 45a = 50 …(2)

by solving (1) and (2), we get u = 175/6 m/s and a = 5/6 m/s^{2}

total distance S travelled before coming rest :- S = (u×u)/(2×a) =

since 500 m was already covered, 10.42 m distance will be travelled before coming to rest