Question

If an object is moving at `8 m/s` and accelerates to `32 m/s` over 6 seconds, what was the object's rate of acceleration?

Answer 1

`4 \ "m/s"^2`

Explanation:

Well, the object changed its velocity by `32 \ "m/s"-8 \ "m/s"` over `6` seconds.

Acceleration is given by the equation:

`a=(s_f-s_i)/t`

where:

• `s_f,s_i` are the final and initial speeds, respectively

• `t` is the time taken

So, we get:

`a=(32 \ "m/s"-8 \ "m/s")/(6 \ "s")`

`=(24 \ "m/s")/(6 \ "s")`

`=4 \ "m/s"^2`

Answer 2

The acceleration is `=4ms^-2`

Explanation:

The initial velocity is `u=8ms^-1`

The final velocity is `v=32ms^-1`

The time is `t=6s`

Apply the equation of motion

`v=u+at`

The acceleration is

`a=(v-u)/t=(32-8)/6=24/6=4ms^-2`

Answer 3

Rate of acceleration `a = 6` m/ `s^2`

Explanation:

`v = u + a t`

`u = 8` m/s, `v = 32` m/s, `t = 6` sec

`a = (v - u) / t = (32 - 8) / 6 = 4` m / `s^2`