Question

A particle moves along the x-axis so that at time t its position is given by s(t) = (t + 3)(t −1)^3, t > 0.

For what values of t is the velocity of the particle decreasing?

0 < t < 1
t > 1
t > 0
The velocity is never decreasing

Answer 1

if s(t) = (t+3)/(t-1)^3
v(t) = ( (t-1)^3 - (t+3)(3)(t-1)^2 )/(t-1)^6 , using the quotient rule
v(t) = (t-1)^2 (t-1 - 3t - 9)/(t-1)^6
= (-2t - 10)/(t-1)^4
= -2(t+5)/(t-1)^4

I will leave it up to you to show that
a(t) = 6(t+7)/(t-1)^5

so the velocity is decreasing for a(t) < 0

looking at 6(t+7)/(t-1)^5
when is this negative?

the critical values are -7 and 1
clearly for t < -7 , a is positive, so no hope there
for t between -7 and 0 , we don't care, since t > 0

let's look between 0 and 1
e.g. t = .5
a(.5) = 6(7.5)/(-.5)^5 , which is negative.
clearly a(1) = 0
and for t > 1, a(t) is positive

looks like 0 < t < 1