A car accelerates from rest at a constant rate a for some time and then decelerates at a constant rate `b` to come to rest. The total time elapsed is `t`.
Total distance travelled by the car is .

Explanation

The velocity time graph for the motion of the car is shown in figure.
Let `t_(1)` be the time taken by the car during its accelerated motion represented by OA part of thegraph. Let `v_(m)` to be the maximum velocity attained by the car after time `t_(1)`.
We know, acceleration = slope of curve OA
` ? a =(AC)/(OC) =(v_(m))/(t_(1))" or "t_(1) =(v_(m))/(a)" ....(i)"`
Let `t_(2)` be the time taken by the car during its retardation motion represented by AB part of the graph. Now, retardation = Slope of curve AB
`p =(AC)/(CB) =(v_(m))/(t_(2))" or "t_(2) =(V_(m))/(p)" ....(ii)"`
Since, total time = T
`? t_(1) +t_(2) =T" or "(v_(m)+v_(m))/(a) = (T)/(p)" or "v_(m) (a+p)/(ap) = T`
or `v_(m) =(ab)/(a+p)" ...(iii)"`
Distance traveled by car, s = area of OAOB
`=(1)/(2) xx" base "xx" height "=(1)/(2) xx T xx v_(m) =(1)/(2) (ap)/(a+p) T^(2)`