An object has a mass of #8 kg#. The object's kinetic energy uniformly changes from #144 KJ# to # 640KJ# over #t in [0, 3 s]#. What is the average speed of the object?

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Answer 1 · 2017-09-18T05:33:06

The average speed is #=307.4ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #=8kg#

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=144000J#

The final kinetic energy is #1/2m u_2^2=640000J#

Therefore,

#u_1^2=2/8*144000=36000m^2s^-2#

and,

#u_2^2=2/8*640000=160000m^2s^-2#

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

The points are #(0,36000)# and #(3,160000)#

The equation of the line is

#v^2-36000=(160000-36000)/3t#

#v^2=41333.3t+36000#

So,

#v=sqrt((41333.3t+36000)#

We need to calculate the average value of #v# over #t in [0,3]#

#(3-0)bar v=int_0^3(sqrt(41333.3t+36000))dt#

#3 barv=[((41333.3t+36000)^(3/2)/(3/2*41333.3))]_0^3#

#=((41333.3*3+36000)^(3/2)/(62000))-((41333.3*0+36000)^(3/2)/(62000))#

#=160000^(3/2)/62000-36000^(3/2)/62000#

#=922.1#

So,

#barv=922.1/3=307.4ms^-1#

The average speed is #=307.4ms^-1#