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.
s(t)=(t+3)(t-1)^3
s ' (t) = (t+3)(3)(t-1)^2 + (t-1)^3 (1)
= (t-1)^2 (3(t+3) + t-1)
= (t-1)^2 (4t + 8)
since we want t > 0 , what values do you think this will give us ?
Remember, for a function to be decreasing the first derivative must
be negative.
we have the velocity, s', above.
To find out when the velocity is increasing, we need its derivative to be positive
s" = 12(t^2-1)
so, as long as t > 1, s' is increasing.
That is, the graph of s(t) is concave up.
To determine when the velocity of the particle is decreasing, we need to find the times t for which the derivative of the position function, s(t), gives a negative value.
First, let's find the velocity function by taking the derivative of the position function with respect to time:
v(t) = derivative of s(t) = d/dt[(t + 3)(t - 1)^3]
To find the derivative, we can expand the position function using the product rule:
v(t) = [(t + 3)(3(t - 1)^2)] + [2(t - 1)^3]
Simplifying further:
v(t) = 3(t + 3)(t - 1)^2 + 2(t - 1)^3
Now, to find when the velocity is decreasing, we need to find when the derivative of the velocity function, v(t), is negative:
dv(t)/dt < 0
Let's take the derivative of v(t) with respect to t:
dv(t)/dt = d/dt[3(t + 3)(t - 1)^2 + 2(t - 1)^3]
Expanding using the product rule:
dv(t)/dt = [3(2(t - 1)(t + 3) + (t + 3)(2(t - 1))] + [6(t - 1)^2]
Simplifying further:
dv(t)/dt = 6(t - 1)(t + 3) + 2(t + 3)(2(t - 1)) + 6(t - 1)^2
Now, let's set this derivative to less than zero and solve for t:
6(t - 1)(t + 3) + 2(t + 3)(2(t - 1)) + 6(t - 1)^2 < 0
Simplifying and factoring:
2(t + 3)[3(t - 1) + 2(t - 1)] + 6(t - 1)^2 < 0
2(t + 3)(5t - 5) + 6(t - 1)^2 < 0
2(t + 3)(5t - 5) + 6(t - 1)(t - 1) < 0
2(t - 1)(5t + 5) + 6(t - 1)(t - 1) < 0
2(t - 1)[(5t + 5) + 3(t - 1)] < 0
2(t - 1)(8t + 2) < 0
Now, to determine the values of t that satisfy this inequality, we can use a number line or consider the signs of each factor.
For (t - 1)(8t + 2) < 0 to be true:
- (t - 1) and (8t + 2) must have opposite signs
- (t - 1) must be negative and (8t + 2) must be positive
or
- (t - 1) must be positive and (8t + 2) must be negative
Solving the factors individually:
(t - 1) > 0 when t > 1
(t - 1) < 0 when t < 1
(8t + 2) > 0 when t > -1/4
(8t + 2) < 0 when t < -1/4
From the above factors, we can conclude that:
- The velocity is decreasing when t < -1/4 and t < 1.
- The velocity is increasing when -1/4 < t < 1.
Since the options provided in the question do not include negative values for t, the correct answer is:
d) The velocity is never decreasing.