An object has a mass of #9 kg#. The object's kinetic energy uniformly changes from #81 KJ# to # 18 KJ# over #t in [0, 4 s]#. What is the average speed of the object?

Question

An object has a mass of #9 kg#. The object's kinetic energy uniformly changes from #81 KJ# to # 18  KJ# over #t in [0,  4  s]#.  What is the average speed of the object?

Amelia Adams · Accepted Answer

The average speed is #=103.0ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #=9kg#

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=81000J#

The final kinetic energy is #1/2m u_2^2=18000J#

Therefore,

#u_1^2=2/9*81000=18000m^2s^-2#

and,

#u_2^2=2/9*18000=4000m^2s^-2#

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

The points are #(0,18000)# and #(4,4000)#

The equation of the line is

#v^2-18000=(4000-18000)/4t#

#v^2=-3500t+18000#

So,

#v=sqrt((-3500t+18000)#

We need to calculate the average value of #v# over #t in [0,4]#

#(4-0)bar v=int_0^4(sqrt(-3500t+18000))dt#

#4 barv=[((-3500t+18000)^(3/2)/(-3/2*3500))]_0^4#

#=((-3500*4+18000)^(3/2)/(-5250))-((-3500*0+18000)^(3/2)/(-5250))#

#=18000^(3/2)/5250-4000^(3/2)/5250#

#=411.8#

So,

#barv=411.8/4=103.0ms^-1#

The average speed is #=103.0ms^-1#

Carson Alexander · Answer

undefined To find the average speed ($v_{\text{avg}}$), use the formula:

$$ v_{\text{avg}} = \frac{\text{Change in distance}}{\text{Change in time}} $$

The kinetic energy change is related to speed ($v$) by the formula:

$$ \text{Change in kinetic energy} = \frac{1}{2} m (v_f^2 - v_i^2) $$

Rearrange to find $v_f$ (final speed):

$$ v_f = \sqrt{\frac{2 \cdot \text{Change in kinetic energy}}{m} + v_i^2} $$

Substitute values and solve for $v_f$, then use it in the average speed formula.