Calculus question

Question

A body travels along a straight line according to the law s=-t^4-4t^3+20t^2,t>=0. At what position after the motion get started,does it come to the rest?

Answer 1

It comes to rest when the velocity is 0. The velocity is the derivative of position.

v = s' = -4t^3 - 12t^2 + 40t = 0

-4t(t^2 + 3t - 10) = 0

-4t = 0

t=0 It's stationary at the start.

t^2 + 3t - 10 = 0

10 = 5 * -2

(t-2)(t+5) = 0

t = 2 and -5

Ignore the - time. t = 2 it's stationary, at rest. Plug that into the original equation to find the position.

-(2^4)-(4*2^3)+20*2^2 = -16 - 32 + 80 = 32 units (whatever units s is in).