Hi, just want to double check my answers and how it is set up! My prof. grades HW very harshly :/

V(t)=t^2-2t-8 1(<=) t (<=) 6

I did this using definite integrals.

Thanks

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- Jul 13th 2010, 07:49 PMsoyeahiknowan easy particle problem
Hi, just want to double check my answers and how it is set up! My prof. grades HW very harshly :/

V(t)=t^2-2t-8 1(<=) t (<=) 6

I did this using definite integrals.

Thanks - Jul 13th 2010, 08:06 PMpickslides
- Jul 14th 2010, 03:59 PMsoyeahiknow
yep! That's what I had it to be, I worked it out and got 66.6 m

but how do I get distance traveled? would I have to take the integral of the displacement? Thanks - Jul 14th 2010, 04:09 PMskeeter
displacement = $\displaystyle \displaystyle \int_{t_1}^{t_2} v(t) \, dt$

distance traveled = $\displaystyle \displaystyle \int_{t_1}^{t_2} |v(t)| \, dt$ - Jul 18th 2010, 11:57 AMsoyeahiknow
my answer for the displacement was 66.6 m

and for the distance traveled, it was 53.3 m

please let me know if I am right. I know that all the signs could have messed me up. Thanks! - Jul 18th 2010, 12:49 PMskeeter
- Jul 18th 2010, 01:11 PMsoyeahiknow
I did the problem wrong. I think I have it now

I know the difference between displacement and distance is that you have the absolute sign for distance. But in this case it doesn't matter since both integrals are positive. Am I right? Thanks

the answer for displacement is 22.3 m

and for distance, it is also 22.3 m - Jul 18th 2010, 01:29 PMskeeter