The position of a particle on the x-axis at time t, t 0, is s(t) = ln(t) with t measured in seconds and s(t) measured in feet. What is the average velocity of the particle for 1 ≤ t ≤ e?

I haven't even the slightest clue as to how I start this question. Please help?

Question

The position of a particle on the x-axis at time t, t > 0, is s(t) = ln(t) with t measured in seconds and s(t) measured in feet. What is the average velocity of the particle for 1 ≤ t ≤ e?

I haven't even the slightest clue as to how I start this question. Please help?

hayisforhorses · Answer

This is pretty simple as long as you have the basic ideas of derivatives and their connection to physics. I would explain more, but I really need to go. This should point you in the right direction.

sydneeod · Answer

well I do but I don't quite know where to start. Is there an equation for average velocity? or is it the same as when someone would just say velocity? I may be overthinking it.

holsteremission · Answer

The particle's average velocity over some time interval is the average rate of change of the particle's position. This is given as the ratio in the change of the position to the change in time:
$$v_\text{average}=\frac{s(b)-s(a)}{b-a}$$

holsteremission · Answer

In this case, $[a,b]=[1,e]$.

sydneeod · Answer

ooooh okay, thank you so much. So from there i would just find v(t), plug in 1 and e, and then plug those into that equation?

holsteremission · Answer

Not quite. There's a difference between average velocity and *instantaneous* velocity. The average velocity is the average rate of change over a given interval, while the instantaneous velocity is the particle's velocity at a specific point in time. The latter can be found by taking a limit of the former (i.e. taking the derivative of the position function) but that's not what's asked here.

sydneeod · Answer

i have to go now, thank you for all the help!

sydneeod · Answer

okay, that makes sense thanks!