An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #144 KJ# to # 120KJ# over #t in [0, 3 s]#. What is the average speed of the object?

2 Answers
Dec 13, 2017

Answer:

Average speed = #(v_0 + v ) /2 approx 256.64 m/s #

Average acceleration = # (v_0 - v ) /3 approx -7.794 m/(s^2) #

Explanation:

For this question we must cosnider the formula for #K.E#

#K.E = 1/2 * m * v^2 #

Initial Kinetic energy#=># #K.E_0= 144*1000 J #

#=> 1/2 * 4 * v_0^2 = 144000#

#=> v_0 approx 268.33 m/s #

Final Kinetic energy, #=> K.E = 120*1000J #

#=> 1/2 * 4 * v^2 = 120000 #

#=> v approx 244.95 m/s #

Average speed = #(v_0 + v ) /2 approx 256.64 m/s #

Average acceleration = # (v_0 - v ) /3 approx -7.794 m/(s^2) #

Dec 13, 2017

Answer:

The average speed is #=256.8ms^-1#

Explanation:

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=4kg#

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=120000J#

Therefore,

#u_1^2=2/4*144000=72000m^2s^-2#

and,

#u_2^2=2/4*120000=60000m^2s^-2#

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

The points are #(0,72000)# and #(3,60000)#

The equation of the line is

#v^2-72000=(60000-72000)/3t#

#v^2=-4000t+72000#

So,

#v=sqrt(-4000t+72000)#

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

#(3-0)bar v=int_0^3(sqrt(-4000t+72000))dt#

#3 barv= (-4000t+72000)^(3/2)/(3/2*-4000)| _( 0) ^ (3) #

#=((-4000*3+72000)^(3/2)/(-6000))-((-4000*0+72000)^(3/2)/(-6000))#

#=72000^(3/2)/6000-60000^(3/2)/6000#

#=770.4#

So,

#barv=770.4/3=256.8ms^-1#

The average speed is #=256.8ms^-1#