The position of a particle on the x-axis at time t, t>0 is ln t. The average velocity....?

Question

The position of a particle on the x-axis at time t, t>0 is ln t. The average velocity of the particle for 1 < or equal to T < or equal to e is ?

Answer 1

The avarage velocit = (Total displacement over the time from 1 to e) / (total time)

total time = e - 1

Total displacement = Integration from 1 to e (lnt dt)
= [tlnt - t] with upper limit e and lower limit 1
= (elne - e - 1ln1 + 1)
= (e - e - 0 + 1) since, lne = 1 and ln1 = 0
= 1

therefore, avarage velocity = 1 / (e - 1)

Answer 2

The average veloctiy is defined as the total distance over the total time. In a time between t=1 and t=e, the position changes from ln(1) to ln(e), so the net distance traveled in this time frame is ln(e) - ln(1) = 1 - 0 = 1. Since the time span is (e-1), the average velocity is 1 / (e-1).

Answer 3

The velocity is dx/dt = 1/t
At t= e is 1/e and at t= 1 is 1
The space berween t=1 and t=e
s=(ln e-ln1) =1 so the average velocity would be 1/(e-1) = space /time

Answer 4

Good question, I was wondering the same thing myself