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Math Help - an easy particle problem

  1. #1
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    an 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
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  2. #2
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    Quote Originally Posted by soyeahiknow View Post
    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
    Did you mean x(t) = \int_1^6 V(t)~dt= \int_1^6 t^2-2t-8~dt ?

    What is your answer?
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  3. #3
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    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
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  4. #4
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    displacement = \displaystyle \int_{t_1}^{t_2} v(t) \, dt

    distance traveled = \displaystyle \int_{t_1}^{t_2} |v(t)| \, dt
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  5. #5
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    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!
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  6. #6
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    Quote Originally Posted by soyeahiknow View Post
    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!
    in any motion problem, distance traveled \ge displacement

    \displaystyle \Delta x = \int_1^6 t^2 - 2t - 8 \, dt

    \displaystyle d = -\int_1^4 t^2 - 2t - 8 \, dt + \int_4^6 t^2 - 2t - 8 \, dt
    Last edited by skeeter; July 18th 2010 at 01:05 PM.
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    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
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  8. #8
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    Quote Originally Posted by soyeahiknow View Post
    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
    no.

    \displaystyle \Delta x = -\frac{10}{3}

    \displaystyle d = \frac{98}{3}
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