An object has a mass of #1 kg#. The object's kinetic energy uniformly changes from #48 KJ# to #36 KJ# over #t in [0, 3 s]#. What is the average speed of the object?

Question

An object has a mass of #1 kg#. The object's kinetic energy uniformly changes from #48 KJ# to #36  KJ# over #t in [0,  3  s]#.  What is the average speed of the object?

Elena Albarran · Accepted Answer

The average speed is #=204.8ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=1kg#

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=48000J#

The final kinetic energy is #1/2m u_2^2=36000J#

Therefore,

#u_1^2=2/1*48000=96000m^2s^-2#

and,

#u_2^2=2/1*36000=72000m^2s^-2#

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

The points are #(0,48000)# and #(3,36000)#

The equation of the line is

#v^2-48000=(36000-48000)/3t#

#v^2=-4000t+48000#

So,

#v=sqrt(-4000t+48000)#

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

#(3-0)bar v=int_0^3(sqrt(-4000t+48000))dt#

#3 barv= [(-4000t+48000)^(3/2)/(3/2*-4000)] _( 0) ^ (3)#

#=((-4000*3+48000)^(3/2)/(-6000))-((-4000*0+48000)^(3/2)/(-6000))#

#=48000^(3/2)/6000-36000^(3/2)/6000#

#=614.3#

So,

#barv=614.3/3=204.8ms^-1#

The average speed is #=204.8ms^-1#

Charles Aeschliman · Answer

undefined To find the average speed, we first need to find the initial and final velocities of the object. We can use the formula for kinetic energy: $ KE = \frac{1}{2} mv^2 $.

Given that the kinetic energy changes uniformly over time, we can use the average kinetic energy: $ KE_{avg} = \frac{1}{2} m(v_{initial} + v_{final}) $.

Using the given kinetic energies and mass, we can solve for the initial and final velocities. Then, we can find the average speed by dividing the total distance traveled by the total time taken.