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.
To find the velocity of the particle, integrate the acceleration.
`=3t^2-18t+v_0` now sub in the initial velocity
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-4)(t-2)|dt` separate the integral to keep positive integrand
`=28/3 approx 9.3`
The total distance travelled is `28/3` .
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