What is the average speed of an object that is moving at #6 m/s# at #t=0# and accelerates at a rate of #a(t) =t^2-5t+3# on #t in [0,3]#?

Question

1338 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-02T06:56:09

The average speed is #=5.25ms^-1#

The speed is the integral of the acceleration

#a(t)=t^2-5t+3#

#v(t)=int(t^2-5t+3)dt#

#=1/3t^3-5/2t^2+3t+C#

Plugging in the initial conditions,

#v(0)=6#

#v(0)=0-0+0+C=6#

#=>#, #C=6#

Therefore,

#v(t)=1/3t^3-5/2t^2+3t+6#

Let the average speed be #barv#

Then,

#(3-0)barv=int_0^3(1/3t^3-5/2t^2+3t+6)dt#

#=[1/12t^4-5/6t^3+3/2t^2+6t]_0^3#

#=(1/12*3^4-5/6*3^3+3/2*3^2+6*3)-(0)#

#=15.75#

#barv=15.75/3=5.25ms^-1#