An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #64 KJ# to # 180 KJ# over #t in [0, 12 s]#. What is the average speed of the object?

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Answer 1 · 2017-06-15T17:37:12

The average speed is #=244.5ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #=4kg#

The initial velocity is #=u_1#

The initial kinetic energy is #1/2m u_1^2=64000J#

The final velocity is #=u_2#

The final kinetic energy is #1/2m u_2^2=180000J#

Therefore,

#u_1^2=2/4*64000=32000m^2s^-2#

and,

#u_2^2=2/4*180000=90000m^2s^-2#

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

The points are #(0,32000)# and #(12,90000)#

The equation of the line is

#v^2-32000=(90000-32000)/12t#

#v^2=4833.3t+32000#

So,

#v=sqrt((4833.3t+32000)#

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

#(12-0)bar v=int_0^12sqrt((4833.3t+32000))dt#

#12 barv=[((4833.3t+32000)^(3/2)/(3/2*4833.3)]_0^12#

#=((4833.3*12+32000)^(3/2)/(7250))-((4833.3*0+32000)^(3/2)/(7250))#

#=90000^(3/2)/7250-32000^(3/2)/7250#

#=2934.6#

So,

#barv=2934.6/12=244.5ms^-1#

The average speed is #=244.5ms^-1#