1. May 12, 2015

### haruspex

2. May 12, 2015

### robphy

Probably the most frequently made error in kinematics is
incorrectly thinking "velocity is distance over time" or,
slightly-better-but-still-bad, "velocity is displacement over time".

Unfortunately, this probably arises from "a common-sense formula" usually encountered via the "average-velocity formula" [which is puzzlingly used to derive the [instantaneous-]velocity formula]---puzzling because a certain-average of a quantity is used to derive the quantity itself.

When misused, the average-velocity of a trip doesn't really help you get the velocity at (say) the endpoint.
If you happen to accelerate uniformly from rest, using "displacement/time" gets you an answer which is "only off by a factor of 2".

.

3. May 12, 2015

### Greg Bernhardt

Nice work haruspex!

4. May 12, 2015

### haruspex

This was in reply to robphy's post#2 in this thread. But I made the reply by clicking on Reply in the comment entry in the chain under the Insights post. This doesn't seem to set the linkage correctly.
Similarly, I replied to DocZaius post#6 that way, but when I looked in this thread it appeared as a reply to a post by another (something like 'ahkron') which I don't see in this thread view at all.
Strange.

Last edited: May 13, 2015
5. May 13, 2015

### robphy

Part of the problem with the typical textbook introduction of "average velocity" is that it often doesn't clarify that
it's a time-weighted average of velocities (which they usually have not yet defined).
So, often the student is often left incorrectly assuming that it's a straight-average of velocities.

For a trip with three piecewise-constant-velocity-legs
\begin{align*} \vec v_{avg} &\equiv \frac{\vec v_1\Delta t_1+\vec v_2\Delta t_2+\vec v_3\Delta t_3}{\Delta t_1+\Delta t_2+\Delta t_3} \\ &= \frac{\Delta \vec x_1+\Delta\vec x_2+\Delta \vec x_3}{\Delta t_1+\Delta t_2+\Delta t_3}\\ &= \frac{\Delta \vec x_{total}}{\Delta t_{total}}\\ \end{align*}

What is usually also missing is an interpretation of the "average velocity".
I use the following...
Can the trip from start position to end position over the same interval of time be done with a constant velocity
(rather than a varying one)? Yes, use the average-velocity over that interval.

Only when the velocity is constant over the interval
will the average-velocity equal the velocity....
that's the only time one can use "[constant] velocity=displacement/time".

For uniform acceleration ,
\begin{align*} \vec v_{avg} &\equiv \frac{\Delta \vec x_{total}}{\Delta t_{total}}\\ &\stackrel{\scriptsize\rm const \ a }{=}\frac{\frac{1}{2}\vec a(\Delta t)^2 +\vec v_i\Delta t}{\Delta t}\\ &\stackrel{\scriptsize\rm const \ a }{=}\frac{1}{2}\vec a(\Delta t) +\vec v_i\\ &\stackrel{\scriptsize\rm const \ a }{=}\frac{1}{2}(\vec v_f -\vec v_i) +\vec v_i \\ &\stackrel{\scriptsize\rm const \ a }{=} \frac{1}{2}(\vec v_f+ \vec v_i) \end{align*}
which looks like a straight-average of the velocities of two piecewise-constant-velocity-legs,
but it's really the straight-average of the starting and ending velocities.

(If this rendered poorly, try reading it here:

6. May 13, 2015

### DocZaius

For Part 3., your final calculation should take the total distance as 2D, not D. Which would make the average speed 3u/2

7. May 13, 2015

### haruspex

Thanks!

8. May 14, 2015

4. Velocity versus acceleration

Another good example of point 4 shows up in simple harmonic motion. When the instantaneous velocity is zero, for example a pendulum at the end point of a swing, the acceleration is a maximum.

9. May 16, 2015

### AbhigyanGoswami

Thanks for the valuable tips.

10. Aug 1, 2015

This Para: Its vertical velocity will be zero after hitting the ground, but the process of hitting the ground involves a large upward acceleration. Applying a SUVAT equation across the period from before to after impact is therefore not valid. The ‘final’ velocity here means at the instant before impact.
is not clear.
The final vertical velocity may not be zero after hitting the ground. It is zero at the instant of hitting the ground, but not zero before and after hitting the ground.

Could you elaborate on the sentence: 'Applying a SUVAT equation across the period from before to after impact is therefore not valid.'