An object has a mass of #3 kg#. The object's kinetic energy uniformly changes from #36 KJ# to #270 KJ# over #t in [0, 9 s]#. What is the average speed of the object?

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Answer 1 · 2017-08-18T17:54:33

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

The kinetic energy is

#KE=1/2mv^2#

The mass is #=3kg#

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=36000J#

The final kinetic energy is #1/2m u_2^2=270000J#

Therefore,

#u_1^2=2/3*36000=24000m^2s^-2#

and,

#u_2^2=2/3*270000=180000m^2s^-2#

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

The points are #(0,24000)# and #(9,180000)#

The equation of the line is

#v^2-24000=(180000-24000)/9t#

#v^2=17333.3t+24000#

So,

#v=sqrt((17333.3t+24000)#

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

#(9-0)bar v=int_0^9(sqrt(17333.3t+24000))dt#

#9 barv=[((17333.3t+24000)^(3/2)/(3/2*17333.3))]_0^9#

#=((17333.3*9+24000)^(3/2)/(26000))-((17333.3*0+24000)^(3/2)/(26000))#

#=180000^(3/2)/26000-24000^(3/2)/26000#

#=2794.2#

So,

#barv=2794.2/9=310.5ms^-1#

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