Question

Need help finishing a calc problem.

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x'(t) is its velocity, and (click link for image) http://tinyurl.com/9hqq7xk is the distance the particle travels in the interval of time.

The position function is given by x(t)=(t-1)(t-3)^2, 0<=t<=5.
Find the total distance the particle travels in 5 units
of time.

Work so far (see attached):

Answer 1

looks fine so far

Answer 2

Yes but where do I go from here

Answer 3

So you want to find the intervals on which the expression in the absolute value is negative, and those on which it is positive. You know that it is a polynomial, so this is easy: we look for the zeroes and then determine the sign of the polynomial between these zeroes (and within the endpoints of integration, of course).

So to find the zeroes we factor:\[3t^2-14t+15=(3t-5)(t-3)\]\[3t^2-14t+15=0\Leftrightarrow t=\frac{5}{3},~t=3.\]You follow me so far?

Answer 4

So we set up the problem so we break up the integration into intervals with the zeroes or something, right?

Answer 5

Yes, something like that...\[3(0)^2-14(0)+15=15,\]so we know that it is positive on the interval (0,5/3). \[3(2)^2-14(2)+15=-1,\]So it is negative on the interval (5/3,3), and positive on the interval (3,5). 5 is the limit of integration, so we don't need to worry about anything else.

Remember, for a negative number a:
\(|a|=-a\)
For a positive number b:
\(|b|=b\)

Can you see what to do next?

Answer 6

My internet bailed out on me but I got it, thanks