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
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
To determine when the velocity of the particle is decreasing, we need to find the intervals where the derivative of the position function, s'(t), is negative.
First, let's find the derivative of the position function, s(t).
s(t) = (t + 3)(t - 1)^3
Using the product rule of differentiation, we have:
s'(t) = (t + 3)(3(t - 1)^2) + (t - 1)^3(1)
Simplifying, we get:
s'(t) = 3(t - 1)^2(t + 3) + (t - 1)^3
To find when the velocity is decreasing, we need to determine when s'(t) is negative.
Let's analyze the algebraic expression to identify the intervals where s'(t) is negative:
1. (t - 1)^2(t + 3): This expression is non-negative for all t since (t - 1)^2 and (t + 3) are both non-negative terms.
2. (t - 1)^3: This expression is also non-negative since it is a cube.
Therefore, we can conclude that s'(t) is never negative, meaning the velocity of the particle is never decreasing.
Therefore, the correct answer is: The velocity is never decreasing.