Confused about kinematic equations

[John3509]

Homework Statement

I am struggling to make them make sense intuitively of the kinematics equations

Relevant Equations

the kinematics equations

x=x₀=vt

---

v=v₀+at

---

x=x₀+v₀t+(1/2)at²



That (1/2) there does not make sense to me. It makes sense mathematically, if you integrate the velocity formula you get the (1/2) as a result of the integration rules and if you differentiate its needed to cancel out the 2 and arrive at the velocity formula

But my line if reasoning does not produce this term, and I can not find the flaw in my reasoning. It is a follows.

When velocity is constant, you get the 1st formula, simple.
But when velocity is not constant, velocity itself is a function of time
so you get

x=x₀+v(t)*t

And the time dependent velocity formula is given in the second equation

v(t)=v₀+at

so plugging it in we get

x=x₀+(v₀+at)t
x=x₀+v₀t+at²

How can I get the (1/2) to show up with this line of thinking? And If I cant then why cant I go down this path?

There is also this other equation I have encountered

v²=v₀²+2aΔx

that I just don't see any intuition behind, why square v in the first place? Any kinematics problem can be solved with the first 3 I listed, so whats up with this one?

 

[Orodruin]

But when velocity is not constant, velocity itself is a function of time
so you get

x=x0+v(t)*t

This is where you are incorrect. Remember that you started out with:

When velocity is constant, you get the 1st formula

In your case, velocity is not constant so you cannot just insert it into a formula that assumes constant velocity.

Instead, you can consider the correct version:
$$
x = x_0 + \bar v(t) t
$$
where $\bar v(t)$ is the average velocity between times 0 and $t$.

If you have constant acceleration $a$ and start with $v(0) = v_0$, what is the average velocity?

 

[Orodruin]

There is also this other equation I have encountered

v²=v₀²+2aΔx

that I just don't see any intuition behind, why square v in the first place? Any kinematics problem can be solved with the first 3 I listed, so whats up with this one?

This is true in the sense that the ones you listed can be used to derive this one (or vice versa). But consider a problem where you are given that an object with constant acceleration starts with velocity $v_0$ and then accelerates to $v$ over a distance $\Delta x$. You are being asked to find the acceleration. Sure, you could start looking for an expression for the time and plug it into another expression with time to derive the above formula, but it is so much simpler if you already know it.

In general, the SUVAT equations are a set of five equations that all describe constant acceleration motion using distance travelled $s$, initial velocity $u$, final velocity $v$, acceleration $a$, and time $t$ (spelling $suvat$). Each of the SUVAT equations relate four of these quantities and can be derived from two of the others by eliminating the unwanted variable. Which one you use depends on what input data you have available.

 

[PeroK]

That (1/2) there does not make sense to me. It makes sense mathematically, if you integrate the velocity formula you get the (1/2) as a result of the integration rules and if you differentiate its needed to cancel out the 2 and arrive at the velocity formula

Displacement is the (signed) area under a velocity vs time graph. For constant acceleration, there is no need to integrate, as the graph is combination of a rectangle and a triangle.

If we take $v_0 = 0$ and $a > 0$, then the graph is a simple triangle. The area of a triangle is half the area of a rectangle with the same base and height. The base is $t$, the height is $at$ and the area is $\frac 1 2 at^2$.

See the good old BBC, for example:

https://www.bbc.co.uk/bitesize/guides/znpp92p/revision/8

 

[John3509]

This is where you are incorrect. Remember that you started out with:


In your case, velocity is not constant so you cannot just insert it into a formula that assumes constant velocity.

Instead, you can consider the correct version:
$$
x = x_0 + \bar v(t) t
$$
where $\bar v(t)$ is the average velocity between times 0 and $t$.

If you have constant acceleration $a$ and start with $v(0) = v_0$, what is the average velocity?

Ok now that makes sense intuitively, but ironically now the arithmetic doesn't make sense to me

you can only use v_av because in a parabola a line from the origin to some point @ (t) is only parallel to a secant line in between the origin and (t). My original set up would only use the velocity at the endpoint (t) which would be too high,

But mathematically, you have two equations, both share the same variable, why cant I just apply a substitution?

x=x₀+vt
v=v₀+at

when a=0
v=v₀

Also, while I understand intuitively why you can place v_av there, mathematically, how is it justified when they are not explicitly the same variable?
I guess what I'm asking here is if you weren't operating with some intuition or sense of conceptual understanding but purely asthmatically how do you arrive at this?

Also, It still doesn't work out does it?

x=x₀+vt
v_av= (v₀+v)/2

How do I arrive at this
x=x₀+v₀t+(1/2)at²
from substitution? Not involving integration.

 

[Orodruin]

But mathematically, you have two equations, both share the same variable, why cant I just apply a substitution?

x=x0+vt
v=v0+at

Because, again, the first formula assumes that v is constant in time and the second explicitly states that it is not (unless a = 0). The two equations therefore describe scenarios that are distinct.

how is it justified when they are not explicitly the same variable?

Effectively, that is the definition of the average velocity and it generally results in
$$
\bar v = \frac 1t \int_0^t v(\tau) d\tau
$$

That is where the expression for average velocity comes from.

Also, It still doesn't work out does it?

Yes it does.

x=x0+vt
vav= (v0+v)/2

How do I arrive at this
x=x0+v0t+(1/2)at2
from substitution? Not involving integration.

Now you can insert v = v0 + at because your $\bar v = (v_0 + v)/2$ assumes constant acceleration, just as v = v0 + at.

 

[kuruman]

But mathematically, you have two equations, both share the same variable, why cant I just apply a substitution?

x=x₀+vt
v=v₀+at

when a=0
v=v₀

All that is true however the two equations that you posted are not applicable to the same physical situation unless a = 0. If the acceleration is not zero the correct way to write the first equation is
$x-x_0=\Delta x=v_{\text{avg.}}\Delta t$
This equation is always applicable whether the acceleration is zero, constant or time-dependent. It allows you to find a number for the average velocity over time interval $\Delta t$ as the distance traveled over that interval divided by the time interval. You can always find a number for that as long as the moving object is at two different positions at two different times.

Now if the acceleration is constant, the average velocity from $t=0$, when the velocity is $v_0$, to $t$, when the velocity is $v$, takes the form $v_{\text{avg.}}=\frac{1}{2}(v_0+v)$ and the distance traveled is $\Delta x=v_{\text{avg.}}\Delta t = \frac{1}{2}(v_0+v)t.$

The two equations to quote are $$\begin{align}& x=x_0+\frac{1}{2}(v_0+v)t \\ &v=v_0+at.\end{align}$$Then if you replace $v$ from equation (2) in equation (1), you get $$x=x_0+v_0t+\frac{1}{2}at^2.$$You may wish to consider tackling the following problem that illustrates the concept of average velocity very nicely.

Problem
An automated speed trap consists of two sensors placed 100 m apart on the side of a straight segment of highway. A car passing the first sensor starts a clock running and stops the clock when it passes the second sensor. The distance of 100 m is divided electronically by the recorded time interval and if the result is more than the speed limit of 25 m/s, a camera is triggered that takes a photo of the license plate for further processing.

A driver enters the speed trap at 45 m/s. What is the highest exit speed that will not trigger the camera assuming that the car keeps moving relative to the highway in the same direction without stopping until it passes the second sensor?

Spoiler

The answer is less than 25 m/s. Why?

Edited to add text in bold red in view of post #8.

 

[Orodruin]

A driver enters the speed trap at 45 m/s. What is the highest exit speed that will not trigger the camera?

c

The car stops at the mid-point for four seconds, then accelerates to a speed arbitrarily close to c

 

[kuruman]

c

The car stops at the mid-point for four seconds, then accelerates to a speed arbitrarily close to c

Right. I edited the problem statement in post #7. It reads more like a legal document now.

 

[Orodruin]

Right. I edited the problem statement in post #7. It reads more like a legal document now.

Still same solution.
What you need to restrict is acceleration to be constant or at least opposite to the direction of motion.

My solution above can easily be adapted to the new criteria above.

 

[nasu]

It doesn't have to stop. Just to go very slowly for a long enough time.
But maybe the point of the problem was to show that the average velocity is not a good indicator of how fast it moved unless the acceleration is known.

 

[kuruman]

But maybe the point of the problem was to show that the average velocity is not a good indicator of how fast it moved unless the acceleration is known.

I think the point of the problem is that there is a difference between average and instantaneous velocity and that measuring the former is not a reliable measurement of the latter. That is why radar guns are used to estimate the instantaneous velocity of objects from cars to tennis balls.

 

[nasu]

Yes, it could be but then the student should be told that the answer (less than 25 km/h) is wrong and why is wrong. There is no such maximum speed unless more information is given.

 

[jbriggs444]

I think the point of the problem is that there is a difference between average and instantaneous velocity and that measuring the former is not a reliable measurement of the latter.

If we assume that position is well defined throughout the time interval and that velocity is well defined throughout the interval except, possibly, at the two endpoints then the mean value theorem requires that the average velocity is a lower bound on the peak instantaneous velocity attained within the open interval.

As has been pointed out, it is not an upper bound on the actual instantaneous velocity at the end of the interval. Or a guarantee that the particle even has a velocity at the end point.

 

[WWGD]

It doesn't have to stop. Just to go very slowly for a long enough time.
But maybe the point of the problem was to show that the average velocity is not a good indicator of how fast it moved unless the acceleration is known.

So, velocity represents the mean and acceleration the standard deviation. Strange how well the analogy (seems to) fit(s).

 

[kuruman]

If we assume that position is well defined throughout the time interval and that velocity is well defined throughout the interval except, possibly, at the two endpoints . . .

What assumptions are appropriate for a car moving on a highway?

 

[robphy]

$\begin{align*}\Delta x&\stackrel{const\ a}{=}v_i\Delta t+ \frac{1}{2}\Delta v\Delta t\\ &\stackrel{const\ a}{=}v_i\Delta t+\frac{1}{2}(a\Delta t)\Delta t\\ \end{align*}$

$\begin{align*}\Delta x &\stackrel{def}{=}v_{avg}\Delta t\\ &\mbox{where v_{avg}\stackrel{const\ a}{=}}\left(\frac{v_i+v_f}{2}\right) \end{align*}$

$\begin{align*}\Delta x &\stackrel{def}{=}v_{avg}\Delta t \stackrel{const\ a}{=} v_{avg} \left(\frac{\Delta v}{a}\right) \end{align*}$

 

[John3509]

If we assume that position is well defined throughout the time interval and that velocity is well defined throughout the interval except, possibly, at the two endpoints then the mean value theorem requires that the average velocity is a lower bound on the peak instantaneous velocity attained within the open interval.

As has been pointed out, it is not an upper bound on the actual instantaneous velocity at the end of the interval. Or a guarantee that the particle even has a velocity at the end point.

Can someone translate this?

 

[John3509]

View attachment 351557$\begin{align*}\Delta x&\stackrel{const\ a}{=}v_i\Delta t+ \frac{1}{2}\Delta v\Delta t\\ &\stackrel{const\ a}{=}v_i\Delta t+\frac{1}{2}(a\Delta t)\Delta t\\ \end{align*}$

View attachment 351558 $\begin{align*}\Delta x &\stackrel{def}{=}v_{avg}\Delta t\\ &\mbox{where v_{avg}\stackrel{const\ a}{=}}\left(\frac{v_i+v_f}{2}\right) \end{align*}$

View attachment 351559$\begin{align*}\Delta x &\stackrel{def}{=}v_{avg}\Delta t \stackrel{const\ a}{=} v_{avg} \left(\frac{\Delta v}{a}\right) \end{align*}$

These are some nice visual aids, what program did you use to make them? My visualization for why v average can be used is a bit different tho.

 

[robphy]

These are some nice visual aids, what program did you use to make them? My visualization for why v average can be used is a bit different tho.View attachment 351935

Thanks.
I used Desmos.

I focused on the velocity-vs-time graph because it’s easier to extract the kinematic formulas from it, compared to the position-vs-time graph.

In addition, my last velocity-vs-time graph gives you motivation for why the velocity appears quadratically in the expression for the displacement.

 

[jbriggs444]

Can someone translate this?

If it takes you one hour to get to a store 60 miles away then your average speed is 60 miles per hour.

This means that your instantaneous speed was exactly 60 miles per hour at some point during the trip. If you started out slow, you must have sped up. If you started out fast, you must have slowed down. In between, your speed must have passed through 60 miles per hour exactly. That's the mean value theorem.

With a few picky and unrealistic exceptions that us mathematicians thought up about continuity, differentiability, open versus closed intervals and such.


The bit about "lower bound" is just math language for "it must be at least as much as this".

Naturally, "upper bound" is the same concept, "it cannot have been more than this".

If the average speed on the way to the store was 60 miles per hour and we know that means that the instantaneous speed was 60 miles per hour at some point then it follows that the maximum speed attained must have been at least 60 miles per hour. So 60 miles per hour would be a "lower bound" on the maximum speed attained.

But it is possible that you went much faster for some brief interval.