Question

If an object with uniform acceleration (or deceleration) has a speed of `2 m/s` at `t=0` and moves a total of 6 m by `t=5`, what was the object's rate of acceleration?

Answer 1

Using `d=V_i*t+1/2at^2` we get `a=-0.32`.

Explanation:

First off, we know acceleration is:
`a=(∆V)/(t)`

We rearrange this to get:
`∆V = at`

Substitute `∆V=V_f-V_i`:
`V_f-V_i=at`

Move `V_i` (initial velocity) over:
`V_f=V_i+at`

Now, we find the displacement by using this formula:
`d=V_i*t + 1/2at^2`

The area beneath a velocity-time graph is the displacement. In this case the area between the line of velocity and the x axis and the interval (0,5) is equal to the displacement or distance traveled over (0,5).

If you have been taught integrals in calculus the integral of velocity is displacement.

This equation is derived from the concept of finding the area beneath the graph. So, plug your numbers in and solve for a:

`6m = 2m`/`s*5s + 1/2a*(5s)^2`
`6m = 10m + 1/2a*25s^2`

Move the 10m over:
`-4m = 1/2a*25s^2`

Multiply both sides by 2 to cancel the 1/2:
`-8m = a*25s^2`

Divide both sides by `25s^2` to get:
`-8/25` `m`/`s^2`
(Which is `-0.32`).

You can check this on a graphing calculator if you wish, by replacing `t` with `x` and inserting `V_i` and `a`.

Hope this helps.

Cheers