Question

If an object with uniform acceleration (or deceleration) has a speed of `12 m/s` at `t=0` and moves a total of 22 m by `t=4`, what was the object's rate of acceleration?

Answer 1

`-3.25"m/s"^2`

Explanation:

Use the equation of motion:

`s=ut+1/2at^2`

`:.22=(12xx4)+(0.5xxaxx4^2)`

`:.22=48+8a`

`:.a=-26/8=-3.25"m/s"^2`