Interval Kinematics

[neilparker62]

For interest - comments welcome.

 

[kuruman]

or interest - comments welcome.

Here are my comments, since you asked.

This approach is, of course, limited to cases where the acceleration is constant. Under constant acceleration, one has to solve the differential equation $\ddot x = \text{const.}$ and get two equations incorporating the initial conditions, $$\dot x(t)=\dot x(0)+(\text{const.})t \tag{1}$$ $$ x(t)=x(0)+\dot x(0)t+\frac{1}{2}(\text{const.})t^2. \tag{2}$$ I think that "rote memorization" should be minimized, not eliminated. We cannot expect students to start from first principles every time they solve a kinematics problem.

In my opinion, the least number of equations to memorize is the above two plus the auxiliary equation obtained by eliminating parameter $t$, $$2(\text{const.})[x(t)-x(0)]=[\dot x(t)^2]-[\dot x(0)]^2 \tag{3} $$ I think that it is important to introduce this equation in kinematics because it will later evolve into mechanical energy conservation.

I am trying to understand why your approach is preferable to the traditional approach. Traditionally, I would say:
Part (a): Find the velocity $u$ at the top of the window.
The known quantities are the transit time $\Delta t=0.1~$s, the transit distance $\Delta h=1.5~$m and the height of the building $H=100~$m. I choose the origin at the top of the window and time $t=0$ when the stone is at the origin. Taking "down" as positive, $x(0)=0$ and $\dot x(0) =u$. Also at $t=\Delta t=0.1~$s, $x(t)=1.5~$m and $\dot x(t)=u$. Equation (2) becomes, $$\begin{align} & x(t)=x(0)+\dot x(0)t+\frac{1}{2}(\text{const.})t^2 \nonumber \\
& \Delta h=u\Delta t+\frac{1}{2}g(\Delta t)^2 \nonumber \\
& \implies u=\frac{2\Delta h-g(\Delta t)^2}{2\Delta t}
=\frac{2\times1.5~\text{m}-9.8~(\text{m/s}^2)\times (0.1~\text{s})^2}{2\times 0.1~\text{s}} = 14.51~\text{m/s}.\nonumber
\end{align}$$Part (b): Find the height of the window.
This part can be answered by using the auxiliary equation as shown in post #1 and will not be repeated here.

So finding $u$ the old fashioned way is just as quick as using the average velocity perspective. The $\pm$ Relationship is obtained directly from Equations (1) and (2) by dividing through by time $t$ which more appropriately should be written as $\Delta t$ to distinguish it from "clock" time $t$. My point here is if you use $t$ for an interval, what symbol will you use for clock time to avoid confusion in students' minds? Clock time becomes important in categories such as "Catch up" problems.

Also, if you insist on using intervals, why is the "drop" relationship limited to $u=0$? Why not use the more general expression and say, $$
\begin{align} & \bar v = \frac{(v+u)}{2}~;~t=\frac{(v-u)}{a}~; ~s=\bar v t \nonumber \\
&\therefore~~s=\frac{(v+u)}{2}\frac{(v-u)}{a}\implies 2as=v^2-u^2 \nonumber
\end{align}$$and then narrow Equation (3) down to $u=0$ if you must?

From my experience, the average velocity concept in constant acceleration problems has limited use in problem-solving but becomes useful in conceptual explanations. A classic example is the speed trap:

The speed limit on a highway is 90 kmh. Two sensors are placed 100 m apart on a straight portion of the highway. When a car passes the first sensor, it triggers a clock and when it passes the second sensor it stops the clock. Software divides the distance (0.1 km) by the measured time interval converted to hours. If the result exceeds "90", a photograph of the license plate is taken and a citation issued.

A speeding motorist notices the trap and starts slowing down but passes the first sensor at 144 kmh. What must his minimum deceleration be (assume it is constant) so that he does not get a ticket?

Most students start solving this problem by assuming that the "safe" exit velocity must be the speed limit of 90 kmh. If you exit at the speed limit, who's to say that you have been speeding, right? Wrong.

They reconsider when they are told that the gadget calculates the average velocity, therefore if the exit velocity is equal to the 90 kmh speed limit, the instantaneous velocity in the 100-meter interval must be always higher than that and the ticket will be issued.

I apologize for saying more than I planned and I stop here.

 

[robphy]

Intuitive: Treats motion as a deviation from the mean.

Did you mean non-inertial motion?


So, with your $a=9.8\rm\ m/s^2$,
upward-increments along the Velocity-axis are associated with "downward acceleration".

Interesting presentation.
I never thought of , for constant-acceleration,
$$v_{i} ,\ v_{f} \quad = v_{avg} \mp (a\Delta t)/2 \quad = \frac{\Delta x}{\Delta t} \mp (a\Delta t)/2 $$

---

In my opinion,

• "velocity" MUST be defined FIRST as the "slope-of-x-vs-t"
• "average-velocity" MUST be defined LATER as the "time-weighted-average of velocities"
  (and that $v_{avg}=(v_f+v_i)/2$ is only for constant-acceleration, as @kuruman suggests)
  
  [My preference is that "velocity" is the first definition students see, otherwise, in my experience, it seems when I ask for "velocity" at certain time [in an accelerated motion, like freefall], I often get (sadly) average-velocity.]
  
  
• "acceleration" MUST be defined as the "slope-of-v-vs-t"
• "displacement" can be interpreted as the "area-under-v-vs-t"
• "average-velocity" should also be interpreted as
  "the [steady] constant-velocity needed to complete the same displacement in the same time-interval",
  which means (as you suggest):
  on a v-vs-t-graph, the height of the rectangle with base $\Delta t$ that has the same area as the area under the v-vs-t graph (here, the area of the trapezoid)

To add more "geometry" to the "Geometry of Motion",
it might be good to point out more geometrical information encoded in the trapezoid,
which could help connect to the standard textbook kinematics equations for constant-acceleration.
See below for details.

For your problem,
using your diagrams,
here is your step 1 (with my emphasis on $(-t/2)$)

here is your step 2 (not to scale)

See my desmos.com/calculator/00069a84c7

 

[Steve4Physics]

Hi @neilparker62, Here are a few additional comments - just in case the presentation will be given again.
______________

Some references to ‘distance’ should be to “displacement”. Similarly for ‘speed’ and ‘velocity’.The distinctions are important.

E.g. in the slide “The Symmetric Equations”:
“s = distance covered during that interval.”
should be
“s = displacement during that interval.”

“The average velocity (s/t) sits exactly halfway between the start and end speeds.”
should be
“The average velocity (s/t) sits exactly halfway between the start and end velocities.”
______________

In the slide entitled ‘The “Drop” Relationship’ you say:
“Since it starts from zero, the average velocity is v/2.
The time taken to reach that velocity is t = v/g.”

The term “that velocity” suggests you are referring to the average velocity, which you are not, So the second sentence should be:
“The time taken to reach the final velocity is t = v/g”
______________

You might want to check your use of significant figures. You have data with precisions of 1 or 2 sig. figs. but give the final answer to 4 sig. figs.
______________

Apologies if I'm being too pedantic!

 

[neilparker62]

Here are my comments, since you asked.

This approach is, of course, limited to cases where the acceleration is constant. Under constant acceleration, one has to solve the differential equation $\ddot x = \text{const.}$ and get two equations incorporating the initial conditions, $$\dot x(t)=\dot x(0)+(\text{const.})t \tag{1}$$ $$ x(t)=x(0)+\dot x(0)t+\frac{1}{2}(\text{const.})t^2. \tag{2}$$ I think that "rote memorization" should be minimized, not eliminated. We cannot expect students to start from first principles every time they solve a kinematics problem.

In my opinion, the least number of equations to memorize is the above two plus the auxiliary equation obtained by eliminating parameter $t$, $$2(\text{const.})[x(t)-x(0)]=[\dot x(t)^2]-[\dot x(0)]^2 \tag{3} $$ I think that it is important to introduce this equation in kinematics because it will later evolve into mechanical energy conservation.

I am trying to understand why your approach is preferable to the traditional approach. Traditionally, I would say:
Part (a): Find the velocity $u$ at the top of the window.
The known quantities are the transit time $\Delta t=0.1~$s, the transit distance $\Delta h=1.5~$m and the height of the building $H=100~$m. I choose the origin at the top of the window and time $t=0$ when the stone is at the origin. Taking "down" as positive, $x(0)=0$ and $\dot x(0) =u$. Also at $t=\Delta t=0.1~$s, $x(t)=1.5~$m and $\dot x(t)=u$. Equation (2) becomes, $$\begin{align} & x(t)=x(0)+\dot x(0)t+\frac{1}{2}(\text{const.})t^2 \nonumber \\
& \Delta h=u\Delta t+\frac{1}{2}g(\Delta t)^2 \nonumber \\
& \implies u=\frac{2\Delta h-g(\Delta t)^2}{2\Delta t}
=\frac{2\times1.5~\text{m}-9.8~(\text{m/s}^2)\times (0.1~\text{s})^2}{2\times 0.1~\text{s}} = 14.51~\text{m/s}.\nonumber
\end{align}$$Part (b): Find the height of the window.
This part can be answered by using the auxiliary equation as shown in post #1 and will not be repeated here.

So finding $u$ the old fashioned way is just as quick as using the average velocity perspective. The $\pm$ Relationship is obtained directly from Equations (1) and (2) by dividing through by time $t$ which more appropriately should be written as $\Delta t$ to distinguish it from "clock" time $t$. My point here is if you use $t$ for an interval, what symbol will you use for clock time to avoid confusion in students' minds? Clock time becomes important in categories such as "Catch up" problems.

Also, if you insist on using intervals, why is the "drop" relationship limited to $u=0$? Why not use the more general expression and say, $$
\begin{align} & \bar v = \frac{(v+u)}{2}~;~t=\frac{(v-u)}{a}~; ~s=\bar v t \nonumber \\
&\therefore~~s=\frac{(v+u)}{2}\frac{(v-u)}{a}\implies 2as=v^2-u^2 \nonumber
\end{align}$$and then narrow Equation (3) down to $u=0$ if you must?

From my experience, the average velocity concept in constant acceleration problems has limited use in problem-solving but becomes useful in conceptual explanations. A classic example is the speed trap:

The speed limit on a highway is 90 kmh. Two sensors are placed 100 m apart on a straight portion of the highway. When a car passes the first sensor, it triggers a clock and when it passes the second sensor it stops the clock. Software divides the distance (0.1 km) by the measured time interval converted to hours. If the result exceeds "90", a photograph of the license plate is taken and a citation issued.

A speeding motorist notices the trap and starts slowing down but passes the first sensor at 144 kmh. What must his minimum deceleration be (assume it is constant) so that he does not get a ticket?

Most students start solving this problem by assuming that the "safe" exit velocity must be the speed limit of 90 kmh. If you exit at the speed limit, who's to say that you have been speeding, right? Wrong.

They reconsider when they are told that the gadget calculates the average velocity, therefore if the exit velocity is equal to the 90 kmh speed limit, the instantaneous velocity in the 100-meter interval must be always higher than that and the ticket will be issued.

I apologize for saying more than I planned and I stop here.

Thanks very much for your detailed comment - no apologies needed. Yes - for this particular lesson on interval kinematics we should indeed be using $\Delta T$ and probably $\Delta v$ and $\Delta s$ as well. I think the use of this equation is indeed conceptual since one can visualize exactly what is happening rather than just use a formula. A moot point since the formula itself (as pointed out by @robphy) can be derived directly from area under the velocity / time graph so can also be 'visualized' although I think not many students will actually appreciate that. Am in the process of developing lesson(s) which will emphasize the geometric / graphical origins of these equations rather than just presenting and using them. Have a few different derivations of the key equation v^2=u^2+2as. (area under v-t curve, conservation of mechanical energy and average velocity x time).

A confession - if it wasn't obvious - is that such lessons (including the above) are "AI-assisted". I come up with the basic lesson idea (eg u,v formula and drop equation above) and AI implements in Latex code. If asked, it can also include graphics (tikz code) and slide presentation (beamer) both via packages available within Latex. So the 'drop equation' in this case is limited to cases where initial velocity = 0 by design. The 'raw' AI lesson can then be modified based on feedback received as I will do here. Minor changes can be directly implemented by Latex code 'tweaks' but mostly AI simply regenerates from scratch.

 

[neilparker62]

Hi @neilparker62, Here are a few additional comments - just in case the presentation will be given again.
______________

Some references to ‘distance’ should be to “displacement”. Similarly for ‘speed’ and ‘velocity’.The distinctions are important.

E.g. in the slide “The Symmetric Equations”:
“s = distance covered during that interval.”
should be
“s = displacement during that interval.”

“The average velocity (s/t) sits exactly halfway between the start and end speeds.”
should be
“The average velocity (s/t) sits exactly halfway between the start and end velocities.”
______________

In the slide entitled ‘The “Drop” Relationship’ you say:
“Since it starts from zero, the average velocity is v/2.
The time taken to reach that velocity is t = v/g.”

The term “that velocity” suggests you are referring to the average velocity, which you are not, So the second sentence should be:
“The time taken to reach the final velocity is t = v/g”
______________

You might want to check your use of significant figures. You have data with precisions of 1 or 2 sig. figs. but give the final answer to 4 sig. figs.
______________

Apologies if I'm being too pedantic!

Not at all. Will try to implement with emphasis on vector rather than scalar quantities. Thanks very much for the feedback. The 'presentation' hasn't been used yet. Am busy developing a set of lessons on kinematics. I thought I would pick that topic that as a 'testing ground' for rapid AI-assisted lesson development. This particular lesson is something of an "aside" from the main 'tree' which focuses on the standard kinematic equations.

 

[neilparker62]

@robphy

Thanks for your kind comment and desmos graph. As mentioned above, I'm working on developing kinematics lessons along the lines you suggest. In addition to the trapezoid area, I also split the area into two triangles leading to the equation s=(u+v)/2 x (u-v)/a.

I also have done some work on showing how the areas work when up is positive and the graph slopes downward with gradient $-9.8 ms^{-2}$

For the particular lesson on interval kinematics and the "drop" equation, I should probably redo the graph so that u-0 = v-u =g (ie passes through the origin). Will have to somehow distinguish between u=0 and v=u for the drop distance calculation against u=u and v=v for the calculation of u and/or v. And - yes - in this instance the graph only makes sense if "down=positive" since it has a positive gradient. The lesson will need to state that explicitly.

 

[robphy]

Instead of initial and final, it might be better to use event-labels: at event A, at time $t_A$, the position is $x_A$, and the velocity is $v_A$, and the acceleration is $a_A$.
Encourage good bookkeeping, especially for multi-stage processes, like your posed problem.