Question

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

Answer 1

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'(t) = a(t) = d/dx (16 - t^2) = -2t`

therefore at point `t = 4,`
`v'(4) = a(4) = -2 * 4 = -8`

therefore , acceleration `= -8`