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?
a) 0<t<1
b) t>1
c) t>0
d) The velocity is never decreasing.
Thanks in advance.
To determine when the velocity of the particle is decreasing, we need to find the derivative of the position function, s(t), with respect to time, which will give us the velocity function, v(t).
To find the derivative of s(t), we can use the product rule and chain rule of differentiation.
s(t) = (t+3)(t-1)^3
Applying the product rule:
s'(t) = (t-1)^3 * d/dt(t+3) + (t+3) * d/dt((t-1)^3)
Applying the chain rule:
s'(t) = (t-1)^3 * 1 + (t+3) * 3(t-1)^2 * d/dt(t-1)
Simplifying:
s'(t) = (t-1)^3 + 3(t+3)(t-1)^2
Now we have the expression for the velocity function, v(t). To find when the velocity is decreasing, we need to find when the derivative of the velocity is negative.
Finding the derivative of v(t):
v'(t) = d/dt[(t-1)^3 + 3(t+3)(t-1)^2]
Expanding and simplifying:
v'(t) = 3(t-1)^2 + 6(t+3)(t-1)
Now, we can solve for when v'(t) < 0 to determine when the velocity is decreasing.
3(t-1)^2 + 6(t+3)(t-1) < 0
Simplifying further:
3(t-1)[(t-1) + 2(t+3)] < 0
3(t-1)(t+1) < 0
To find the values of t that satisfy this inequality, we can set each factor equal to zero and find when the expression changes sign:
t - 1 = 0 --> t = 1
t + 1 = 0 --> t = -1
Now we can test different intervals of t using a number line to see when the expression is negative (indicating the velocity is decreasing):
For t < -1:
If we choose t = -2, the expression becomes positive:
3(-2-1)(-2+1) > 0
For -1 < t < 1:
If we choose t = 0, the expression becomes negative:
3(0-1)(0+1) < 0
For t > 1:
If we choose t = 2, the expression becomes positive:
3(2-1)(2+1) > 0
From the tests on the intervals, we can conclude that the velocity is negative (decreasing) for -1 < t < 1.
Therefore, the answer is:
a) 0 < t < 1