An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #24 KJ# to # 42KJ# over #t in [0, 9 s]#. What is the average speed of the object?

2 Answers
Mar 31, 2017

Answer:

#4 m/s#

Explanation:

Let the rate of increase of kinetic energy with time be some #m (KJ)/s# where #m# is a constant as given in the question.

#therefore (dE(t))/(dt) = m# where #E # is Kinetic Energy and #t# is time.

#implies dE(t)=m*dt#
#implies ##intdE = int m*dt#

#implies# #E(t) = m*t + c# where #c# is a constant of integration.
We have two conditions:-
1. #E(0) = 24# &
2. #E(9) = 42#

from 1. & 2.
#24 = c# & # 42 = 9*m + 24 implies m=2 #
#therefore E(t) = 2*t + 24#

#implies (1/2)*m*v^2 = 2*t+24# where #m# is the mass & #v# is the velocity (speed) of body.

#implies v^2 = t+12#
#implies v = sqrt(t+12)#

Average speed = #(int_(t_1)^(t_2)v*dt)/(int_(t_1)^(t_2)dt)#
where
# t_1 = 0# & #t_2=9#

#therefore# Avg. speed #= (int_(0)^(9)sqrt(t+12)*dt)/(int_(0)^(9)dt)#
#= (2/3*(t+12)^(3/2))/(t)]_0^9 = 2/3*(21sqrt21 - 12sqrt12)/9 approx 4 #

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Mar 31, 2017

Answer:

#128.05 ms^-1#, rounded to two decimal places.

Explanation:

Let rate of change of kinetic energy be
# (dKE(t))/(dt) = C#
where #C # is constant with time #t#.

Integrating both sides with #t# we get
#int (dKE(t))/(dt) cdot dt = int C cdotdt#

#=>KE= Ccdott + c# ......(1)
where #c# is a constant of integration.

To evaluate #c#, use initial condition at #t=0#
#24xx10^3 = Ccdot0+c#
#=>c=24xx10^3#
We have expression for kinetic energy as
#KE= Ccdott + 24xx10^3#

To evaluate #C#, use final condition at #t=9#
#42 xx10^3= Ccdot9+24xx10^3#
#=>C=2xx10^3#

Equation (1) becomes
#KE= (2cdott + 24)xx10^3# ......(2)

If #m# is the mass and #v(t)# is the velocity of body, Kinetic energy is given as
#KE=1/2mv^2#
#=> 1/2mv^2(t) = (2*t+24)xx10^3#
Given is #m=4kg#. Above expression becomes
#=> 1/2xx4v^2(t) = (2*t+24)xx10^3#
#=> v(t)^2 = (t+12)xx10^3#
# => v= +-sqrt((t+12)xx10^3)#
# => "speed "|v|= sqrt((t+12)xx10^3)#

Average speed #="Total Distance traveled"/"Time taken"=(int_0^9|v(t)|*dt)/(9-0)#

#:.# Average speed #= (int_(0)^(9)sqrt((t+12)xx10^3)*dt)/9#
#= sqrt(10^3)/9[(t+12)^(3/2)/(3/2)]_0^9 #
#= (2sqrt(10^3))/27(21^(3/2) - 12^(3/2))#
#= 128.05 ms^-1#, rounded to two decimal places.