A car moves in a straight line, the car accelerates from rest with a constant acceleraation `alpha` on a straight foad. After gaining a velocity `v`, the car moves with that velocity for somerime. Then car decelerates with a retardation `beta`, If the total distance covered by the car is equal to `s` find the total time of its motion.

Answer

`(s)/(v)+(v)/(2) (1 (alpha) +(1)/(beta)`

Explanation

Let the car accelerate for a time `t_(1)`. Move uniformly for a time `t_(2)` and then retard for a time `t_(3)`as shown in the `v-t` graph.
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Then the total time `T` of the car is
`T=t_(1)+t_(2)+t_(3)` (i)
The slope of `v-t` graph
gives `(v)/(t_(1))=alpha` and `(-v)/(t_(3)) =-beta`
Then `t_(1)+t_(3) =(v)/(beta)` (ii)
Area under `v-t` graph gives the total displacement
s=Area of the trapezinum `=(v)/(2)(t_(2)+T)`
This gives `t_(2)=(2s)/(v)-T` (iii)
Substituting `t_(1)+t_(3)` from Eq. (ii) and `t_(2)` From Eq. (iii) in Eq. (i) we
have `T=(v)/(alpha)+(v)/(beta)-(2s)/(v)-T`
This yields `T=(s)/(v)+(v)/(2)((1)/(alpha)+(1)/(beta))`.