Question

If an object with uniform acceleration (or deceleration) has a speed of `4 m/s` at `t=0` and moves a total of 26 m by `t=9`, what was the object's rate of acceleration?

Answer 1

`gamma=-20/81`

Explanation:

The covered space for an uniform accelerated movement is given by
`s = s_0+v_0t+1/2 gamma t²`
In `t=0` the movement begins with `s_0` and `v_0=4`
After `t=9` the object situation is
`26 = 4 * 9 + 1/2 gamma 9^2`. Solving for `gamma=-20/81`