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

Let 'u' be the initiz-l speed of the rrctcrcycce when it covers the frSlr150 m in 10 s.LE! 'z> be the accelerenon of the rr ororcycte. The
velocity a~er this 10 second is,
v = u- at
=>v=u+l0a
Now, using,
5=ur- V2C1t)
=> 2503= lOu - YI.(10 ')
=> 250 = 10 u + 50,
=c- u + 5, = 25 "" ",,{l)
For the next 20 s the. Inltia~veloery IS 'V (= t.; - lOa).
Usil)Q. 5 = vt/ - '/2 at i'
=) 250 = (U - 10a)20+ !13(20 })
.) 250 = 20u • 200.1- 200d
.) 250 = 20u - 400.1
.> 25 • 2u + 40.
-> 2" - 40.- 25 """"",,,,,,,,,,,(2)
(2),2'(1) -, 2" - 40. - 2" - 10. - 25 - 50
-> ;0. - ,25
.> a. ·25/30. ·5/6 fn/S i
(1)-> u- Sa- 2S
->u-25 - 5a-25+2S/6'175/6mjs
Let vlbe tile velocity after the 20 s.
vl·v·~t/
-> v'·(u-l0.) +3(20)
... vl• u- 30-3
Lor. SI be the distance travelled after the 20 s be'ore oomin9 to rest,
a •
-) 0 - (u + 30~)1+2.51
.) O. (175/6 - ;0-5/6)', 2'{5/6)S i
=,0 =(175/6·25)' - (5!3)SI
=> (5/3) S 1= (25/6) )
=> (Si3}S/= 625/35
=> Sl, 125/12 m