At the instant, the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of `3.20 ms^(-2)`, At the some instant, a truck travelling with a constant speed of `20.0 ms^(-1)`, overtakes and passes the car.
a. At what distance from its starting point does the car overtake the truck?
b. Calculate the speed of the car when it overtakes the truck.
c. Sketch an `x-t` graph of themotion of both vehicles.
Take `x-0` at the intersection.
d. Sketch a `v_(x)-t` graph of the motion of both vehicles.

Answer

`250 m`; `40.0 m s^(-1)`

Explanation

The truck's position as a function of time is given by `x_(T)=v_(T) t`, with `v_(t)` being the truck's constant speed and the car's position is given by ` x_(C)=(1//2)a_(C)t^(2)`. Equating the two expressions anddividing by a factor of `t` (this reflects the fact that the car and the truck are at the same place at `t=0` and solving for `t` yields
`t=(2v_(T))/(a_(C)) =(2(20.0 ms^(-1)))/(3.20 ms^(-2)) =12.5 s`
and at this time `x_(T)=x_(C)=250 m`
b.`v_(c)=a_(C)t=(3.20 ms^(-2)) (12.5 s) =40.0 ms^-1)`
c. .
d. Let us see when the velocities of both become same `S4. 39`. For this
`v_(c)=v_(t) rArr a_(c)t=v_(t) rArr 3.2t=20 rArr t=6.25 s`
.