An object has a mass of #9 kg#. The object's kinetic energy uniformly changes from #99 KJ# to # 45 KJ# over #t in [0, 4 s]#. What is the average speed of the object?

Answer 1

The average speed is #=125.7ms^-1#

The kinetic energy is

#KE=1/2mv^2#
mass is #=9kg#
The initial velocity is #=u_1#
#1/2m u_1^2=99000J#
The final velocity is #=u_2#
#1/2m u_2^2=45000J#

Therefore,

#u_1^2=2/9*99000=22000m^2s^-2#

and,

#u_2^2=2/9*45000=10000m^2s^-2#
The graph of #v^2=f(t)# is a straight line
The points are #(0,22000)# and #(4,10000)#

The equation of the line is

#v^2-22000=(10000-22000)/4t#
#v^2=-3000t+22000#

So,

#v=sqrt((-3000t+22000)#
We need to calculate the average value of #v# over #t in [0,4]#
#(4-0)bar v=int_0^4sqrt((-3000t+22000))dt#
#4 barv=[((-3000t+22000)^(3/2)/(-3/2*3000)]_0^4#
#=((-3000*4+22000)^(3/2)/(-4500))-((-3000*0+22000)^(3/2)/(-4500))#
#=22000^(3/2)/4500-10000^(3/2)/4500#
#=502.9#

So,

#barv=502.9/4=125.7ms^-1#
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Answer 2

To find the average speed, we need more information, such as the initial and final velocities of the object. Without this information, the average speed cannot be determined.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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