A motor car starts from rest and accelerates uniformly for `10 s` to a velocity of `30 ms^(-1)`. It then runs at a constant speed and is finally brought to distance covered is `830 m`. Find the value of acceleration, retardation and total time taken.

Explanation

Let `S_(1), S_(2) and S_(3)` be the distances covered by motor boat with uniform acceleration, constant velcity and uniform retardation respectively.
Taking motion car while moving with auniform accleration, we have
`u=0, t_(1) =10 s, v=30 ms^(-1) , a=?`
As, `v=u +at_(1), so 30 =0 +a xx 10`
or `a=3 ms^(-2)`
Distance coverd, `S_(1) =ut +1/2 at_(1)^(2)`
`=0 +1/2 xx 3 xx 10^(2) =15 m`
Taking motor car while moving with uniform velcity, we have
`S_(1) =150 ,: A_(3) =80 m` and S=830 m`
As `S_(1) + S_(2) +S_(3) =830`
:. 150 +S_(2) +80 =830`
or `S_(2) =830 -230 =600 m`
This distance is coverd with a uniform velcity
`30 ms^(-1)`, so time taken, `t_(2) (600)/(30) =20 s`
Taking moton of motor car with uniform retardation, we have
`u=30 ms^(-1), v=0, S=80m, a=?`
As `v^(2) =u^(2) +2 aS So, o= 30^(2) +2 a xx 80`
or `a =- (900)/(160) =-(45)/8 ms^(-1) =- 5. 625 ms^(-2)`
Time taken, `t_(3) =(v-u)/a =(0-(30))/(-45//8) =5 33 s`
Total time taken `=t_(1) +t_(2) +t_(3)`
`=10 +20 +5.33`
`=35.33 s`.