Question

If an object with uniform acceleration (or deceleration) has a speed of `5 m/s` at `t=0` and moves a total of 15 m by `t=12`, what was the object's rate of acceleration?

Answer 1

Any acceleration problem can be solved given three variables and the kinematics equations.

Explanation:

In kinematics problems, there are five variables:
`vec v_i " "` (initial velocity)
`vec v_f " "` (final velocity)
`vec a " "` (acceleration)
`Delta vec d " "` (displacement)
`Delta t " "` (elapsed time)

and five equations, each of which only uses four of the variables:

`Delta vec d = vec v_i t + 1/2vec a Delta t^2`
`Delta vec d = vec v_f t - 1/2vec a Delta t^2`
`Delta vec d = (vec v_i + vec v_f)/2 Delta t`
`vec a = (vec v_f - vec v_i)/(Delta t)`
`vec v_f^2 = vec v_i2 + 2 vec a Delta vec d`

so to solve any question, we need three variables, and the equation that has the fourth we are trying to find. In this question, we have our initial velocity, displacement, and time, and we are looking for acceleration. The equation that has those four variables is:

`Delta vec d = vec v_i t + 1/2vec a Delta t^2`

so let's insert our values and solve for the unknown:

`15m = 5m/cancel(s)*12cancel(s) + 1/2vec a(12s)^2`
`15m=60m + 1/2*144s^2*vec a`
`15m-60m=72s^2*vec a`
`(-45m)/(72s^2)=vec a`
`vec a = -0.625m/s^2`

This process works for any acceleration problem where we have three of the five variables, and are looking for a fourth.