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What does "Just before" and "Just after" really mean in physics problems?

Physics Asked by Normal_Vector on May 10, 2021

So I’m stuck in a dynamics problem that asks what is the acceleration of a body just after A, where A is the point that separates the motion of the body from a curvilinear path to projectile motion. May I ask what acceleration should I use? Should I use the $x$-component of acceleration from curvilinear, or the $0$ $x$-component of acceleration from projectile? Also, what should I use for the $y$-component acceleration? Is it $g$?

3 Answers

You can treat "just before" and "just after" as left-hand limits and right-hand limits respectively.

If you want the behavior of some quantity that is a function of time $f(t)$, then $f$ right before and right after $t=t_0$ would be $$f_text{before}=lim_{tto t_0^-}f(t)$$ $$f_text{after}=lim_{tto t_0^+}f(t)$$

Other answers go into more detail into how you would obtain $f(t)$ if system parameters relevant to $f$ change at $t=t_0$. But if you already have $f(t)$ for all times (or all valid times needed for these limits) then you should be good to go from here.

Answered by BioPhysicist on May 10, 2021

Although, I cannot comment on the specifics of your problem (due to the homework policy and the lack of details in the question), I can make a general statement of what it means by "before" and "after".

In physics you are asked to create mathematics models that correspond to particular situations. Sometimes these situations undergo a defined change in behavior, which necessitates a change in the mathematical model.

Find how something behaves just after such change is much simpler than "a lot later". You can look at acceleration (derivative of velocity) right after the event with a limit that is biased on one side only.

$$ begin{aligned} a^{rm after} & = lim_{delta rightarrow 0} frac{ v(t_0+delta) - v(t_0) }{delta} = left. frac{{rm d}v}{{rm d} t} right|_{t>t_0} a^{rm before} & = lim_{delta rightarrow 0} frac{ v(t_0) - v(t_0-delta) }{delta} = left. frac{{rm d}v}{{rm d} t} right|_{t<t_0} end{aligned}$$

Mathematically all you have to do a choose the correct model that describes the behavior before or after the event that changed the physics of the problem.

Simplified Explanation

The slope of $f(x)=|x|$ is $+1$ after zero and $-1$ before zero. The zero point changes the math of the problem, and we are looking at the behavior either before or after the change.

Answered by JAlex on May 10, 2021

I am sure that you have met this idea before but not noticed it.

For example when you are investigating the collision of two objects you most often deal with before the collision and after collision but never even consider what is happening during the collision and the time during which the collision occurs.
In such an example the approximation is made that the collision time is much, much less than any other times involved.

To describe such a situation one might label the time just before the collision $t=0^-$, the time of the collision $t=0$ and the time just after the collision $t=0^+$.
These are just labels to differentiate between the three phases of the motion.

So once you are in the realm of the time just after A you should use parameters which a pertinent to the time after the event which happened at A and in your example that is anything to do with the projectile motion.

Answered by Farcher on May 10, 2021

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