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
Jan 1, 2018

Answer:

#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#