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
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
#v(t)=int(t-8)dt#
#=>v(t)=1/2t^2-8t+C#
We can use that
#18=1/2(0)^2-8(0)+C#
#=>C=18#
Therefore, we have:
Average velocity is given by:
#v_"avg"=(Deltas)/(Deltat)#
We can take the integral of
#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.
