What is the average speed of an object that is moving at #18 m/s# at #t=0# and accelerates at a rate of #a(t) =t-8# on #t in [1,4]#?

1 Answer
Aug 14, 2017

Answer:

#v_"avg"=1.5" m"//"s"#

Explanation:

We are given the following information:

  • #|->abs(vecv)=18" m"//"s at " t=0#

  • #|->a(t)=t-8#

  • #|->t in [1,4]#

We can begin by recognizing that acceleration is the derivative of velocity, or equivalently, velocity is the antiderivative/integral of acceleration.

We can take the integral of the provided equation for acceleration to find the equation for velocity at time #t#.

#v(t)=int(t-8)dt#

#=>v(t)=1/2t^2-8t+C#

We can use that #v=18# when #t=0# to find the constant #C#:

#18=1/2(0)^2-8(0)+C#

#=>C=18#

Therefore, we have: #v(t)=1/2v^2-8t+18#

Average velocity is given by:

#v_"avg"=(Deltas)/(Deltat)#

We can take the integral of #v(t), t in [1,4]# to find the position #Deltas#.

#Deltas=int_1^4(1/2t^2-8t+18)dt#

#=>=1/6t^3-4t^2+18t|_1^4#

#=>1/6(4)^3-4(4)^2+18(4)-1/6(1)^3+4(1)^2-18(1)#

#=>64/6-64+72-1/6+4-18#

#=>9/2 " m"#

Substituting this into our equation for average velocity:

#v_"avg"=(9/2" m")/((4-1)"s")#

#=>=(9/2" m")/(3" s")#

#=>color(darkblue)(3/2" m"//"s")#

#=>=1.5" m"//"s"#

Speed is the magnitude of velocity, and so the answer remains the same.