a(t) = 1- ta(t)=1−t
implies v(t) = t- t^2/2 + alpha⇒v(t)=t−t22+α
v(0) = 0 implies alpha = 0v(0)=0⇒α=0
So: v(t) = t- t^2/2 v(t)=t−t22
implies x(t) = t^2/2- t^3/6 + alpha ⇒x(t)=t22−t36+α
x(0) = 0 implies alpha = 0x(0)=0⇒α=0
implies x(t) = t^2/2- t^3/6 ⇒x(t)=t22−t36
v_(ave)|_0^1 = (int_0^1 v(t) dt)/(1 - 0)vave∣10=∫10v(t)dt1−0
= (x(1) - x(0))/(1) = 1/3 " m/s"=x(1)−x(0)1=13 m/s
Now here's where you have to be careful. That's the average velocity (vv), not speed (ss). And s(t) = abs (v(t))s(t)=|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)v(t)=t−t22=t(1−t2).
So the particle stops at t = 0, 2t=0,2
And for t in [0,1], v(t) ge 0 implies s = abs v = vt∈[0,1],v(t)≥0⇒s=|v|=v
So, in this case, the average speed is also the average velocity.
