Total Distance of a Velocity Function

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To find the total distance traveled by an object moving along the x-axis from the initial position x(0)=3 over the interval [0,3], the key is to integrate the absolute value of the velocity function v(t) = 4 - t^2. The integral from 0 to 3 gives the net displacement, but since the velocity changes sign within this interval, the total distance must account for this change. The correct approach involves calculating the integral of the absolute value of the velocity to ensure all movement is counted positively. The misunderstanding stemmed from not recognizing the importance of the sign change in the velocity function.
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Homework Statement


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].


Homework 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.


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.
 
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2ltben said:

Homework Statement


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].


Homework 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.

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)?
 
Stupid of me to forget the function.
v(t) = 4-t^2
 
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)
 
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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