An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #18 KJ# to # 48KJ# over #t in [0, 9 s]#. What is the average speed of the object?

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Answer 1 · 2017-10-28T06:36:30

The average speed is #=127.3ms^-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=18000J#

The final kinetic energy is #1/2m u_2^2=48000J#

Therefore,

#u_1^2=2/4*18000=9000m^2s^-2#

and,

#u_2^2=2/4*48000=24000m^2s^-2#

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

The points are #(0,9000)# and #(9,24000)#

The equation of the line is

#v^2-9000=(24000-9000)/9t#

#v^2=1666.7t+9000#

So,

#v=sqrt((1666.7t+9000)#

We need to calculate the average value of #v# over #t in [0,9]#

#(9-0)bar v=int_0^12(sqrt(1666.7t+9000))dt#

#9 barv=[((1666.7t+9000)^(3/2)/(3/2*1666.7))]_0^9#

#=((1666.7*9+9000)^(3/2)/(2500))-((-1666.7*0+9000)^(3/2)/(2500))#

#=24000^(3/2)/2500-9000^(3/2)/2500#

#=1145.7#

So,

#barv=1145.7/9=127.3ms^-1#

The average speed is #=127.3ms^-1#