The graph below represents the velocity `vs.` time graph for a car.

(A) During which intervals is the car at rest?
(B) During which intervals is the car moving a constant nonzero velocity ? State the velocity at each of these intervals
(C) There are two intervals during which the car is accelerating. Find the acceleration in each of these intervals.
(D) Sketch the acceleration `vs`. time garph that corresponds to the motion of the car form `0s` to `14s`.

Explanation

(A) The car is at rest when the velocity is zero.: `A` to `B` and `E` to `F`.
(B) The velocity of the car is constant form `C` to `D`, when its velocity is `+10m//s`, and from `G` to `H`,when its velocity is `-10m//s`. The car is raveling backward during the interval `G` to `H` compared to the interval `C` to `D`.
(C) The car is accelerating during intervals `B` to `C` and `F` to `G`. During the interval `B` to `C`, the acceleration can be found be finding the slope of the `v vs,.t` graph:
slope of a `v vs.t` graph `= (rise)/(run) = (Deltav)/(Deltat) = (10m//s-0)/(6s-2s)`
`=+-2.5ms^(2)`.
During the interval `F` to `G`:
slope `= (rise)/(run) = (Deltav)/(Deltat) = (-10m//s-0)/(12s-10s) =- 5m//s^(2)`
In this last case, the car is still speeding up (from `0`to `-10m//s^(2))`, it's just speeding a negative direction
(D) The acceleration `vs`. time graph looks like this