An object has a mass of #3 kg#. The object's kinetic energy uniformly changes from #48 KJ# to # 32KJ# over #t in [0,12s]#. What is the average speed of the object?

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Answer 1 · 2017-09-23T05:46:04

The average speed is #=163.0ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=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=48000J#

The final kinetic energy is #1/2m u_2^2=32000J#

Therefore,

#u_1^2=2/3*48000=32000m^2s^-2#

and,

#u_2^2=2/3*32000=21333.3m^2s^-2#

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

The points are #(0,32000)# and #(12,21333.3)#

The equation of the line is

#v^2-32000=(21333.3-32000)/12t#

#v^2=-888.9t+32000#

So,

#v=sqrt((-888.9t+32000)#

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

#(12-0)bar v=int_0^12(sqrt(-888.9t+32000))dt#

#12 barv=[((-888.9t+32000)^(3/2)/(-3/2*888.9))]_0^12#

#=((-888.9*12+32000)^(3/2)/(-1333.3))-((-888.9*0+32000)^(3/2)/(-1333.3))#

#=32000^(3/2)/1333.3-21333.3^(3/2)/1333.3#

#=1956.4#

So,

#barv=1956.4/12=163.0ms^-1#

The average speed is #=163.0ms^-1#