An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #96 KJ# to # 270 KJ# over #t in [0, 12 s]#. What is the average speed of the object?

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Answer 1 · 2017-05-05T11:26:11

The average speed is #=299.5ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=4kg#

The initial velocity is #=u_1#

#1/2m u_1^2=96000J#

The final velocity is #=u_2#

#1/2m u_2^2=270000J#

Therefore,

#u_1^2=2/4*96000=48000m^2s^-2#

and,

#u_2^2=2/4*270000=135000m^2s^-2#

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

The points are #(0,48000)# and #(12,135000)#

The equation of the line is

#v^2-48000=(135000-48000)/12t#

#v^2=7250t+48000#

So,

#v=sqrt((7250t+48000)#

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

#(12-0)bar v=int_0^12sqrt((7250t+48000))dt#

#12 barv=[((7250t+48000)^(3/2)/(3/2*7250)]_0^12#

#=((7250*12+48000)^(3/2)/(10875))-((7250*0+48000)^(3/2)/(10875))#

#=135000^(3/2)/10875-48000^(3/2)/10875#

#=3594.1#

So,

#barv=3594.1/12=299.5ms^-1#