An object has a mass of #3 kg#. The object's kinetic energy uniformly changes from #96 KJ# to #240 KJ# over #t in [0, 9 s]#. What is the average speed of the object?

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Answer 1 · 2017-05-13T10:19:57

The average speed is #=332.01ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=3kg#

The initial velocity is #=u_1#

#1/2m u_1^2=96000J#

The final velocity is #=u_2#

#1/2m u_2^2=240000J#

Therefore,

#u_1^2=2/3*96000=64000m^2s^-2#

and,

#u_2^2=2/3*240000=160000m^2s^-2#

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

The points are #(0,64000)# and #(9,160000)#

The equation of the line is

#v^2-64000=(160000-64000)/9t#

#v^2=10666.67t+64000#

So,

#v=sqrt((10666.67t+64000)#

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

#(9-0)bar v=int_0^9sqrt((10666.67t+64000))dt#

#9 barv=[((10666.67t+64000)^(3/2)/(3/2*10666.67)]_0^9#

#=((10666.67*9+64000)^(3/2)/(16000))-((10666.67*0+64000)^(3/2)/(16000))#

#=160000^(3/2)/16000-64000^(3/2)/16000#

#=2988.07#

So,

#barv=2988.07/9=332.01ms^-1#