An object has a mass of #3 kg#. The object's kinetic energy uniformly changes from #75 KJ# to #45 KJ# over #t in [0, 9 s]#. What is the average speed of the object?

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Answer 1 · 2017-06-26T06:24:26

The average speed is #=199.5ms^-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=75000J#

The final kinetic energy is #1/2m u_2^2=45000J#

Therefore,

#u_1^2=2/3*75000=50000m^2s^-2#

and,

#u_2^2=2/3*45000=30000m^2s^-2#

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

The points are #(0,50000)# and #(9,30000)#

The equation of the line is

#v^2-50000=(30000-50000)/9t#

#v^2=-2222.2t+50000#

So,

#v=sqrt((-2222.2t+50000)#

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

#(9-0)bar v=int_0^3sqrt(-2222.2t+50000))dt#

#9 barv=[((-2222.2t+50000)^(3/2)/(-3/2*2222.2)]_0^9#

#=((-2222.2*9+50000)^(3/2)/(-3333.3))-((-2222.2*0+50000)^(3/2)/(-3333.3))#

#=50000^(3/2)/3333.3-30000^(3/2)/3333.3#

#=1795.3#

So,

#barv=1795.3/9=199.5ms^-1#

The average speed is #=199.5ms^-1#