# Deceleration of a point when it is momentarily at rest moving along a straight line with velocity 16-t^2?

Feb 13, 2018

acceleration $= - 8$

#### Explanation:

the point at which it is at rest is the point when its velocity is 0 so we can find that as $16 - {t}^{2} = 0$ to find that $t = 4$ when it is at rest (Note that time only moves forward and starts from 0 so $t = - 4$ is not an option)

the derivative of velocity is acceleration, therefore $v ' \left(t\right) = a \left(t\right) = \frac{d}{\mathrm{dx}} \left(16 - {t}^{2}\right) = - 2 t$

therefore at point $t = 4 ,$
$v ' \left(4\right) = a \left(4\right) = - 2 \cdot 4 = - 8$

therefore , acceleration $= - 8$