An object has a mass of #9 kg#. The object's kinetic energy uniformly changes from #135 KJ# to # 36KJ# over #t in [0, 4 s]#. What is the average speed of the object?

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Answer 1 · 2017-03-14T11:25:12

The average speed is #=135.8ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The initial velocity is #=u_1#

#1/2m u_1^2=135000J#

The final velocity is #=u_2#

#1/2m u_2^2=36000J#

Therefore,

#u_1^2=2/9*135000=30000m^2s^-2#

and,

#u_2^2=2/9*36000=8000m^2s^-2#

The graph of #v^2=f(t)# is a straight line

The points are #(0,30000)# and #(4,8000)#

The equation of the line is

#v^2-30000=(8000-30000)/4t#

#v^2=-5500t+30000#

So,

#v=sqrt((-5500t+30000)#

We need to calculate the average value of #v# over #t in [0,4]#

#(4-0)bar v=int_0^5sqrt(-5500t+30000) dt#

#4 barv=[(-5500t+30000)^(3/2)/(3/2*-5500)]_0^4#

#=((-5500*4+30000)^(3/2)/(-8250))-((-5500*0+30000)^(3/2)/(-8250))#

#=8000^(3/2)/-8250-30000^(3/2)/-8250#

#=1/8250(30000^(3/2)-8000^(3/2)))#

#=543.1#

So,

#barv=543.1/4=135.8ms^-1#