An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #42 KJ# to # 320 KJ# over #t in [0, 12 s]#. What is the average speed of the object?

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Answer 1 · 2017-05-06T09:13:08

The average speed is #=292.4ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=4kg#

The initial velocity is #=u_1#

#1/2m u_1^2=42000J#

The final velocity is #=u_2#

#1/2m u_2^2=320000J#

Therefore,

#u_1^2=2/4*42000=21000m^2s^-2#

and,

#u_2^2=2/4*320000=160000m^2s^-2#

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

The points are #(0,21000)# and #(12,160000)#

The equation of the line is

#v^2-21000=(160000-21000)/12t#

#v^2=11583.3t+21000#

So,

#v=sqrt((11583.3t+21000)#

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

#(12-0)bar v=int_0^12sqrt((11583.3t+21000))dt#

#12 barv=[((11583.3t+21000)^(3/2)/(3/2*11583.3)]_0^12#

#=((11583.3*12+21000)^(3/2)/(17375))-((1583.3*0+21000)^(3/2)/(17375))#

#=160000^(3/2)/17375-21000^(3/2)/17375#

#=3508.3#

So,

#barv=3508.3/12=292.4ms^-1#