What is the average speed of an object that is still at #t=0# and accelerates at a rate of #a(t) =2t^2-3t-2# from #t in [2, 4]#?

1 Answer
Dec 27, 2015

Answer:

#15.32"m/s"#

Explanation:

Acceleration is a function of time here so the average velocity will be the area under the acceleration - time graph between the limits given.

graph{2x^2-3x-2 [-10, 10, -5, 5]}
#:.v_(ave)=int_2^4a.dt#

#v_(ave)=int_2^4(2t^2-3t-2).dt#

#v_(ave)=[(2t^3)/3-(3t^2)/2-2t]_(2)^(4)#

#v_(ave)=[(2xx4^(3))/(3)-(3xx4^(2))/2-(2xx4)]-[(2xx2^(3)-(3xx2^(2))/2-(2xx2)]#

#v_(ave)=[128/3-24-8]-[16/3-6-4]#

#v_(ave)=10.66-(-4.66)#

#v_(ave)=15.32"m/s"#