# Please check my work on this short calc (?) prob

1. Sep 4, 2007

### frasifrasi

1. The problem statement, all variables and given/known data
Two particles move along an x axis. Position of particle 1 is x = 6t^2 + 3t + 2. The acceleration of particle 2 is given by a = -8t and at t=0, its velocity is 20 m/s. When the velocities of the particles match, what is their velocity?

2. Relevant equations

3. The attempt at a solution

Ok, so I took the derivative of the position function and found it to be v = 12t + 3. Then I took the integral from the acceleration function and found it to be v = -4t^2 + C ( by plugging t=0, I got C = 20). Hence, v = -4t^2 + 20.

Now, I equated both velocity functions and was left with 4t^2 + 12t -17. Does that mean I should solve the quadratic to get two answers--this seems too simple and the problem is supposed to be 3/3 for difficulty...

2. Sep 4, 2007

### rootX

yes, that's what I did also, before reading yours work.

3. Sep 4, 2007

### Staff: Mentor

Your work looks right to me.

4. Sep 4, 2007

### frasifrasi

Thank you, but how is it that you come out with two aswers? Should I eliminate the negative one or is it physically possible in this case?

5. Sep 4, 2007

### rootX

because it's quadratic v = -4t^2 + 20

why you hate negative value lol?
how's it different from positive?