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) =t8# on #t in [1,4]#?
1 Answer
Explanation:
We are given the following information:

#>abs(vecv)=18" m"//"s at " t=0# 
#>a(t)=t8# 
#>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(t8)dt#
#=>v(t)=1/2t^28t+C#
We can use that
#18=1/2(0)^28(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^28t+18)dt#
#=>=1/6t^34t^2+18t_1^4#
#=>1/6(4)^34(4)^2+18(4)1/6(1)^3+4(1)^218(1)#
#=>64/664+721/6+418#
#=>9/2 " m"#
Substituting this into our equation for average velocity:
#v_"avg"=(9/2" m")/((41)"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.