An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #48 KJ# to # 360 KJ# over #t in [0, 12 s]#. What is the average speed of the object?

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Answer 1 · 2017-06-21T07:02:10

The average speed is #=310.5ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #=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=48000J#

The final kinetic energy is #1/2m u_2^2=360000J#

Therefore,

#u_1^2=2/4*48000=24000m^2s^-2#

and,

#u_2^2=2/4*360000=180000m^2s^-2#

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

The points are #(0,24000)# and #(12,180000)#

The equation of the line is

#v^2-24000=(180000-24000)/12t#

#v^2=13000t+24000#

So,

#v=sqrt((13000t+24000)#

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

#(12-0)bar v=int_0^12sqrt((13000t+24000))dt#

#12 barv=[((13000t+24000)^(3/2)/(3/2*13000)]_0^12#

#=((13000*12+24000)^(3/2)/(19500))-((13000*0+24000)^(3/2)/(19500))#

#=180000^(3/2)/19500-24000^(3/2)/19500#

#=3725.6#

So,

#barv=3725.6/12=310.5ms^-1#

The average speed is #=310.5ms^-1#