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Math Help - Given velocity, find the distance traveled?

  1. #1
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    Given velocity, find the distance traveled?

    I would really appreciate some help with the following:

    The velocity function is v(t)= -t^2 +6t -8. Find the distance traveled during the time interval [0,6].

    I know that the formula for distance traveled = |v(t)|*dt on the given time interval [a,b], so my formula would be |-t^2+6t-8| where a=0 and b=6.

    I calculated the anti-derivate of v(t) to be -(t^3)/3 +3t^2 -8t.

    I was able to calculate the displacement to be -12, but I don't think that has any relevance to this particular problem?

    At this point, I'm absolutely stuck. Any explanations of the process of this computation would be very greatly appreciated.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Fawx View Post
    I would really appreciate some help with the following:

    The velocity function is v(t)= -t^2 +6t -8. Find the distance traveled during the time interval [0,6].

    I know that the formula for distance traveled = |v(t)|*dt on the given time interval [a,b], so my formula would be |-t^2+6t-8| where a=0 and b=6.

    I calculated the anti-derivate of v(t) to be -(t^3)/3 +3t^2 -8t.

    I was able to calculate the displacement to be -12, but I don't think that has any relevance to this particular problem?

    At this point, I'm absolutely stuck. Any explanations of the process of this computation would be very greatly appreciated.
    No the displacement is different from the distance (similar to the way velocity is different from speed). The displacement is -12 meters.

    To calculate the distance, you need to take the absolute value of the velocity function, like you said. On the interval [0,6], the function is negative on [0,2] and [4,6] and positive on [2,4].

    So your distance is given by \left|\int_0^2 -t^2+6t-8\,dt\right|+\int_2^4-t^2+6t-8\,dt + \left|\int_4^6 -t^2+6t-8\,dt\right|

    You can easily integrate all of these and find that the distance traveled is \frac{20}{3}+\frac{4}{3}+\frac{20}{3} = \frac{44}{3} meters.

    (You will also see that if you take -\frac{20}{3}+\frac{4}{3}-\frac{20}{3} you will get -12, which is exactly what you calculated for displacement.)
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  3. #3
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    Thank you so much; that definitely helped me tie it all together.
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