What is the average speed of an object that is still at t=0 and accelerates at a rate of a(t) = 36-t^2 from t in [2, 6]?

1 Answer
Andrew I.
Jan 1, 2018

117.3ms^-1

Explanation:

To get the average speed, we must sum the velocity and divide by the time traveled. We have a continuous function so we must integrate to sum the velocity. So average velocity is given by:

v_(avg) = 1/(Delta t)int_(t_i)^(t_f)v(t)dt

Now, we have been given the acceleration which means we must first integrate to find the formula for velocity, so:

v(t) = inta(t)dt = int 36-t^2 dt=36t-1/3t^3+C

Using the fact that the object is still at at t=0 then:

v(0) = 36(0)-1/3(0)^3+C=0 -> C=0

So we have for the velocity:

v(t) = 36t-1/3t^3

Now we can use this to get the average velocity, in our case the initial time t_i=2s and the final time t_f=6s so Delta t = 6-2=4s

Now putting that into the expression for the average velocity:

1/(Delta t)int_(t_i)^(t_f) v(t)dt = 1/4int_2^6 36t-1/3t^3dt

=1/4[18t^2-1/12t^4]_2^6

=1/4({18(6)^2-1/12(6)^4}-{18(2)^2-1/12(2)^4})

=1408/12~~117.3ms^-1

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