Acceleration/velocity/Displacement problem

**abclarinetuvwxyz** · January 6th 2009, 03:56 PM

The acceleration of an object is given by the function a(t) = -t/2 + 9/4. Also at time t=0, the velocity of the object is -2. Find the difference between the distance and the displacement traveled by the object to the nearest integer from t=0 to t=10.

Okay so this is what i have so far...

I took the integral of the acceleration to get the velocity and what I obtained was 9t/4 - (t^2)/4 + C
then I got -2 = C

I then integrated the velocity equation to get the position equation and I got -(t^3)/12 + (9t^2)/8 -2t + C.

How do I find C for the position equation?
Or did I go to far...

I know I need to set an equation equal to zero to determine min and max to see where the function changes direction and then I set up an integral of the same function from 0 to one point then from that point to 10...

can someone please help me?

**skeeter** · January 6th 2009, 04:36 PM

The acceleration of an object is given by the function a(t) = -t/2 + 9/4. Also at time t=0, the velocity of the object is -2. Find the difference between the distance and the displacement traveled by the object to the nearest integer from t=0 to t=10.

displacement ... $\Delta x = \int_0^{10} v(t) \, dt$

displacement is a straight-forward FTC calculation.

however, for distance traveled ... $d = \int_0^{10} |v(t)| \, dt$

$v(t) = \frac{-t^2 + 9t - 8}{4} = -\frac{(t-8)(t-1)}{4}$

$v(t) = 0$ at $t = 1$ and $t = 8$

$\int_0^{10} |v(t)| dt = -\int_0^1 v(t) \, dt + \int_1^8 v(t) \, dt - \int_8^{10} v(t) \, dt$

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