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

1 Answer
May 26, 2017

Answer:

#-0.5"m"/"s"#

Explanation:

Since average speed is given by the equation

#v_("av-x") = (Deltax)/(Deltat)#

We need to find the positions of the car at times #t = 2# and #t = 3#.

We can derive first the velocity equation from the acceleration, and then the position from that velocity equation, via integration, and we'll assume the initial position and velocity are #0#:

#v_x(t) = int_0^t a_xdt = 2/3t^3 - t^2 -2t#

#x(t) = int_0^t v_xdt = 1/6t^4 - 1/3t^3 - t^2#

Now we have our position equation, #x(t)#, so let's plug in the values #2# and #3# for #t# to find the position at those times:

#x(2) = 1/6(2)^4 - 1/3(2)^3 - (2)^2 = -4.0"m"#

#x(3) = 1/6(3)^4 - 1/3(3)^3 - (3)^2 = -4.5"m"#

So, the average velocity of the car on the interval #t in [2,3]# is

#v_("av-x") = ((-4.5 "m") - (-4.0 "m"))/(3"s" - 2"s") = color(red)(-0.5"m"/"s"#