Acceleration/ total distance

**CalcGeek31** · January 5th 2009, 01:13 PM

Hi I was starting a total distance problem and got stuck.

The acceleration of an object is given by a(t) =6sin(t) with initial velocity of -9.5. Find the total distance the object travels on the interval [0, pi] to the nearest integer.

I know to take the integral and solve for velocity. I got the equation v(t) = -6cos(t) - 3.5. Firstly, is that right? and secondly How do I continue from there?

**skeeter** · January 5th 2009, 01:22 PM

velocity function is correct

total distance traveled = $\int_0^{\pi} |v(t)| \, dt$

**CalcGeek31** · January 5th 2009, 01:27 PM

Then am I right in saying that 11 would be the correct answer?

**skeeter** · January 5th 2009, 01:39 PM

no ... not the distance traveled.

the displacement is -11, but in this case, displacement $\neq$ distance traveled

**CalcGeek31** · January 5th 2009, 01:43 PM

Oh so I would have to determine where the object changes direction and add the area from both curves?

**skeeter** · January 5th 2009, 01:48 PM

total distance = $-\int_0^a v(t) \, dt + \int_a^{\pi} v(t) \, dt$

where $a = \arccos\left(-\frac{7}{12}\right)$

**CalcGeek31** · January 5th 2009, 01:53 PM

Okay thank you so much, I have been having problems with that one for a while.

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