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The acceleration of a particle moving in a straight line is given by a(t) = 6(t-3).   Given that particle’s initial velocity is 24 units per second, during the time interval 0 less than or equal to t is less than or equal to 5, find (c) The total distance travelled by the particle.

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To find the velocity of the particle, integrate the acceleration.

`v=6int(t-3)dt`

`=3t^2-18t+v_0`   now sub in the initial velocity

`=3t^2-18t+24`

Now to get the total distance from 0 to 5, we need the integral of the absolute value of the velocity from 0 to 5.  This means:

`d=int_0^5|3t^2-18t+24|dt`   factor the integrand

`=3int_0^5|t^2-6t+8|dt`

`=3int_0^5|(t-4)(t-2)|dt`   separate the integral to keep positive integrand

`=3(int_0^2(t^2-6t+8)dt+int_2^4(-t^2+6t-8)dt+int_4^5(t^2-6t+8)dt)`

`=3((1/3t^3-3t^2+8t)_0^2+(-1/3t^3+3t^2-8t)_2^4+(1/3t^3-3t^2+8t)_4^5)`

`=28/3 approx 9.3`

The total distance travelled is `28/3` .

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