Question

If an object with uniform acceleration (or deceleration) has a speed of `3 m/s` at `t=0` and moves a total of 25 m by `t=8`, what was the object's rate of acceleration?

Answer 1

`0.03m/s^2`

Explanation:

Recall the following kinematic formula of motion:

`color(blue)(|bar(ul(color(white)(a/a)Deltad=v_iDeltat+1/2aDeltat^2color(white)(a/a)|)))`

where:
`Deltad=`change in distance
`v_i=`initial velocity
`Deltat=`change in time
`a=`acceleration

Start by listing out the given and required values.

`v_i=3m/s`

`Deltad=25m`

`t=8s`

`a=?`

Notice how the only variable to solve for is `a`.

Thus, rearrange the formula for `a` and plug in your known values into the formula to find the acceleration of the object.

`Deltad=v_iDeltat+1/2aDeltat^2`

`=1/2aDeltat^2=Deltad-v_iDeltat`

`=(2(Deltad-v_iDeltat))/(Deltat^2)`

`=(2(25m-(3m/s(8s))))/(8s)^2`

`=0.03125m/s^2`

`~~color(green)(|bar(ul(color(white)(a/a)0.03m/s^2color(white)(a/a)|)))`