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The velocity function is v(t)= -(t^2)+6t-8for a particle moving along a line.   Find the displacement and the distance traveled by the particle during the time interval [-2,5].

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lfryerda eNotes educator | Certified Educator

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The displacement for the particle is the integral of the velocity between the two time endpoints.  This means that 

`d=int_{-2}^5(-t^2+6t-8)dt`   use power laws




`=-112/3 approx -37.3`

The total displacement is `-112/3` .

To find the total distance, we need to look at the integral of the absolute value of the velocity between the bounds.  This means that the distance is:



`=int_{-2}^5|(t-2)(t-4)|dt`   need to find where the integrand is positive

`=int_{-2}^2|t^2-6t+8|dt+int_2^4|-t^2+6t-8|dt+int_4^5|t^2-6t+8|dt`  now integrate





`=120/3 approx 140.3`

The total distance is `120/3` .

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