An object has a mass of #3 kg#. The object's kinetic energy uniformly changes from #75 KJ# to #48 KJ# over #t in [0, 9 s]#. What is the average speed of the object?

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Answer 1 · 2017-05-05T11:36:40

The average speed is #=202.1ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=3kg#

The initial velocity is #=u_1#

#1/2m u_1^2=75000J#

The final velocity is #=u_2#

#1/2m u_2^2=48000J#

Therefore,

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

and,

#u_2^2=2/3*48000=32000m^2s^-2#

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

The points are #(0,50000)# and #(9,32000)#

The equation of the line is

#v^2-50000=(32000-50000)/9t#

#v^2=-2000t+50000#

So,

#v=sqrt((-2000t+50000)#

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

#(9-0)bar v=int_0^9sqrt((-2000t+50000))dt#

#9 barv=[((-2000t+50000)^(3/2)/(-3/2*2000)]_0^9#

#=((-2000*9+50000)^(3/2)/(-3000))-((-2000*0+50000)^(3/2)/(-3000))#

#=50000^(3/2)/3000-32000^(3/2)/3000#

#=1818.7#

So,

#barv=1818.7/9=202.1ms^-1#