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

1 Answer
Feb 13, 2018

acceleration = -8 =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 16t2=0 to find that t = 4t=4 when it is at rest (Note that time only moves forward and starts from 0 so t = -4t=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