# If an object with uniform acceleration (or deceleration) has a speed of 8 ms^-1 at t=0 and moves a total of 16 m by t=7, what was the object's rate of acceleration?

Feb 13, 2017

We can use the formula $s = u t + \frac{1}{2} a {t}^{2}$, and rearrange to make $a$ the subject:

$a = \frac{2 \left(s - u t\right)}{t} ^ 2 = \frac{2 \left(16 - 7 \times 8\right)}{7} ^ 2 = \frac{2 \left(- 40\right)}{49} = - 1.63 m {s}^{-} 2$

#### Explanation:

The total time taken is $t = 7$ $s$.

(Note that it's sometimes a little confusing: 't = 7' might describe a moment in time, 7 seconds after the start, or it might describe a period of time. In this case it's being used one way in the question and a different way in the answer.)

If the object had continued at the same velocity, $8$ $m {s}^{-} 1$, for $7$ $s$, it would have covered $56$ $m$. As it was, it only covered $16$ $m$. From a common sense perspective, this means it must have slowed down (decelerated), so the negative number in our answer when we calculated the acceleration of the object makes perfect sense.