# Total Distance of a Velocity Function

1. May 8, 2007

### 2ltben

1. The problem statement, all variables and given/known data
An object moves along the x-axis starting off from the initial position x(0)=3.
What is the total distance traveled by the object over the time interval [0,3].

2. Relevant equations
The integral of the function from 0 to a plus the absolute value of the integral of the function from a to 3, where F(a) = 0.

3. The attempt at a solution
I've tried a number of attempts to factor out any zeroes and, finding none, I resorted to a TI-89, confirming the assumption(not good, seeing as this is an AP Cal AB no-calculator practice test, for a test which is tomorrow morning).

I don't know where I'm going wrong in my procedure, but I keep getting 3. The answer I know is 23/3, and the initial position shouldn't matter a bit since its a definite integral and will only be negated anyway. It should be a simple integral from 0 to 3 of the velocity function with respect to t, but I keep getting the wrong answer.

2. May 9, 2007

### siddharth

Can you post the question exactly as it was given to you? Is the velocity of the particle as a function of time given in the question? What's F(a)?

3. May 9, 2007

### 2ltben

Stupid of me to forget the function.
v(t) = 4-t^2

4. May 9, 2007

### Office_Shredder

Staff Emeritus
The integral from zero to three of the velocity will give you the displacement, which is total net distance from the starting point.

So if you double back on yourself, your distance will continue to increase, but your displacement will decrease. So what you want is the integral of the absolute value of the velocity (note it changes signs once between zero and three)

5. May 9, 2007

### 2ltben

My problem was with the sign change I guess. I refused to touch my graphing calculator and didn't think to graph the function by hand.