What is the average speed of an object that is still at #t=0# and accelerates at a rate of #a(t) =1-t# from #t in [0, 1]#?

1 Answer
Apr 28, 2017

#a(t) = 1- t#

#implies v(t) = t- t^2/2 + alpha#

#v(0) = 0 implies alpha = 0#

So: #v(t) = t- t^2/2 #

#implies x(t) = t^2/2- t^3/6 + alpha #

#x(0) = 0 implies alpha = 0#

#implies x(t) = t^2/2- t^3/6 #

#v_(ave)|_0^1 = (int_0^1 v(t) dt)/(1 - 0)#

# = (x(1) - x(0))/(1) = 1/3 " m/s"#

Now here's where you have to be careful. That's the average velocity (#v#), not speed (#s#). And #s(t) = abs (v(t))#. So negative velocities count as positive speeds. We need therefore to be sure of the direction of travel of the object.

We note:

#v(t) = t- t^2/2 = t(1 - t/2)#.

So the particle stops at #t = 0, 2#

And for #t in [0,1], v(t) ge 0 implies s = abs v = v#

So, in this case, the average speed is also the average velocity.