An electric vehicle starts from rest and accelerates at a rate of `2.0 m//s^(2)` in a straight line until it reaches a speed of 20 m//s. The vehicle then slows at a constant rate of `1.0m//s^(2)` until it stops. (a) How much time elapses from start to stop ? (b) How far does the vehicle travel from start to stop ?

Explanation

KEY IDEA
We separate the motion into two parts, and take the direction of motion to be positive. In part 1, the vehicle accelerates from rest to its highest speed, we are given `v_(0)= 0`, `v=20 m//s` and `a=2.0 m//s^(2)`. In part 2, the vechicle decelerates from its highest speed to a halt, we are given `v_(0)=20` m/s, v = 0 and `a = - 10 m//s^(2)` ( negative because the acceleration vector points opposite to the direction of motion).
Calculations : (a) From Table 2-1, we find `t_(1)` ( the duration of part 1) from `v =v_(0)+a t`. In this way, `20=0+2.0t_(1)` yields `t_(1)=10s`. We obtain the duration `t_(2)` of part 2 from the same equation . Thus, `0=20 + (-1.0)t_(2)` leads to `t_(2)=20s`, and the total is `t = t_(1)+t_(2)=30s`.
(b) For part 1, taking `x_(0)=0`, we use the equation `v^(2)=v_(0)^(2)+2a(x-x_(0))` from Table `2-1` and find
`x = v^(2)-v_(0)^(2)+2a(x-x_(0))` from Table 2-1 and find
`x=(v^(2)-v_(0)^(2))/(2a)= ((20 m//s)^(2)-(0)^(2))/(2(2.0 m//s^(2)))=100m`.
This position is then the initial position for part 2, so that when the same equaiton is used in part 2 we obtain
`x-100 m = (v^(2)-v_(0)^(2))/(2a)= ((0)^(2)-(20 m//s)^(2))/(2(-1.0 m//s^(2)))`.
Thus, the final position is `x=300m`. That this is also the total distance traveled should be evident ( the vehicle did not reverse its direction of motion ).