An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #128 KJ# to # 36 KJ# over #t in [0, 12 s]#. What is the average speed of the object?

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Answer 1 · 2017-11-14T07:15:40

The average speed is #=200ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=4kg#

The initial velocity is #=u_1ms^-1#

The final velocity is #=u_2 ms^-1#

The initial kinetic energy is #1/2m u_1^2=128000J#

The final kinetic energy is #1/2m u_2^2=36000J#

Therefore,

#u_1^2=2/4*128000=64000m^2s^-2#

and,

#u_2^2=2/4*36000=18000m^2s^-2#

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

The points are #(0,64000)# and #(12,18000)#

The equation of the line is

#v^2-64000=(18000-64000)/12t#

#v^2=-3833.3t+64000#

So,

#v=sqrt(-3833.3t+64000)#

We need to calculate the average value of #v# over #t in [0,12]#

#(12-0)bar v=int_0^12(sqrt(-3833.3t+64000))dt#

#12 barv=[((-3833.3t+64000)^(3/2)/(-3/2*3833.3))]_0^12#

#=((-3833.3*12+64000)^(3/2)/(-5750))-((-3833.3*0+64000)^(3/2)/(-5750))#

#=64000^(3/2)/5750-18000^(3/2)/5750#

#=2395.8#

So,

#barv=2395.8/12=200ms^-1#

The average speed is #=200ms^-1#