# 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].

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

`=(-1/3t^3+3t^2-8t)_{-2}^5`

`=-125/3+75-40-(8/3+12+16)`

`=-133/3+7`

`=-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:

`d=int_{-2}^5|-t^2+6t-8|dt`

`=int_{-2}^5|t^2-6t+8|dt`

`=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

`=(1/3t^3-3t^2+8t)_{-2}^2+(-1/3t^3+3t^2-8t)_2^4+(1/3t^3-3t^2+8t)_4^5`

`=8/3-12+16-(-8/3-12-16)-64/3+48-32-(-8/3+12-16)`

`+125/3-75+40-(64/3-12+32)`

`=112/3+4/3+4/3`

`=120/3 approx 140.3`

The total distance is `120/3` .

Approved by eNotes Editorial Team