What is the average speed, on #t in [0,5]#, of an object that is moving at #12 m/s# at #t=0# and accelerates at a rate of #a(t) =5t^2-3t# on #t in [0,4]#?

1 Answer
Feb 12, 2018

Given:
Initial velocity of the object #v0=12m/s# at #t=0#
To determine the average speed for# t(0,5)#
The acceleration is given by
#a(t)=5t^2-3t#
We know that
#a(t)=(dv)/dt#
Separating variables,
#dv=a(t)dt#
Substituting for #a(t)#
#dv=(5t^2-3t)dt#
Integrating both sides
#intdv=int(5t^2-3t)dt#
#v=5/3t^3-3/2t^2+v0#

At #t=0, v0=12m/s#
Substituting
#12=5/3(0)^3-3/2(0)^2+v0#
#12=0-0+v0#
#v0=12#
Thus, the expression for the velocity given by
#v(t)=5/3t^3-3/2t^2+12#
Velocity at 4 sec is
#v(4)=5/3(4)^3-3/2(4)^2+12#
#v(4)=106.667-37.5+12#
#v(4)=81.167#m/s
Average velocity for the interval is
Distance travelled is obtained by integrating the expression of velocity
#v=(ds)/dt#
Separating variables,
#ds=v(t)dt#
Substituting for #v(t)#
#dv=5/3t^3-3/2t^2+12dt#

Integrating both sides
#intdv=int(5/3t^3-3/2t^2+12)dt#
#s(t)=5/12t^4-1/2t^3+12t+s0#
Assuming the object starts from rest, s0=0
#s(t)=5/12t^4-1/2t^3+12t+s0#
At t-4s, the position is
#s(4)=5/12(4)^4-1/2(4)^3+12t+0#
#s(4)=122.667#

Total distance travelled in the interval of 4 seconds is #122.667#m
Total time elapsed is 4 seconds

average speed = distance /time
#=122.667/4=30.667 m/s#