A car starts form rest and accelerates uniformly for `10 s` to a velocity of `8 ms ^(-1)`. It then runs at a constant velocity an dis finally brought to rest in `64 m` with a constant retardation. The total distance covered by the car is `584 m` Find the value of acceleration, retardation, and total time taken.

Explanation

`8=0+a_(1)xx10 rArr a_(1) =0.8 m s^(-2)`
`0^(2)=8^(2)+2a_(2)xx64 rArr a_(2) =-0.5 ms^(-2)`
`0=8+a_(2)t_(3) rArr t_(3)=16 s`
.
`S_(1)=(1)/(2) a_(1)t^(2)=(1)/(2) (0.8) 10^(2) =40 m`
`S_(2)=584-40-64=480 m, t_(2)=s_(2)/(v) =(480)/(8) =60 s`
Total time taken `=10 +60=86 s`.