How do you guys deal with problems having only 2 variable? Good fomula sheet?

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In summary, Rick is asking how many equations are there that are common in kinematics problems, and how to remember them. He is also worried about problems in high school physics and whether or not he will be able to do well on the upcoming math test. The first equation he mentions is t=v-vnaught/a, which is a derivative of s=v-vnaught. Rick explains that all constant-acceleration problems can be solved using these two equations, and that if he practices using them, he shouldn't have to worry about the other equations. The last sentence of the summary is that if you want to go the calculus route, the only equation you ever really need is a = constant!
  • #1
Torec_Scrail
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Hey guys, sorry this is kinda vauge (Can't sleep) How do you guys deal with problems listing only 2 variables, so you can't straight use the 5 main velocity motion formulas? I know some like t=v-vnaught/a
But, how many all are there that are common? I've ran into 2*vnaught/a
But I'm not used to them, and can't quite see always when to use em..
Any tips or stuff I can read on the net on these? And what formulas would you guys put down on a formula sheet? I want to commit those two to memory (I have, but I might forget em in a weeK)
Anyways..I'll try to check on this tomorrow and maybe expand..thx.
 
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  • #2
Well, I'm not sure. If you knew more math, then once you know the relation between s, v, and a, then you can derive anything. However, I'm assuming you are in high school and you probably don't have the neat math tricks that you will learn later on in your education.

Anyhoo, one golden equation that really helped me in high school physics was:
[tex] s = v_{o}t + \frac{1}{2}at^{2} [/tex]
Most kinematic problems you do in high school will use them and also in your first year university physics.

I don't know what else to say, but that's the start!

Cheers,
-Rick
 
  • #3
Oh, I was good at HS Physics, I even help a friend out with his a bit..but, the uni ones are def more complicated..
Well this 1 isn't that complicated..but, worried like freeze up in the test totally forget the velocity is 0
(a) with what speed mst a ball be thrown vertically from ground level to rise to a maximum height of 50m? (b) how long will it be in the air?
31m and 6.4 s

I guess it's kinda like the 1 I helped friend with..ball stays in air for 8 seconds being shot straight up, what was its intial velocity? I just took half and basically saw how fast it hit the ground from it's highest point..

But, like, your eqn...I don't see how some gets t = 2(v initial)/a from it..
the t's kinda cancel so its not square rooted?

I'm not sure how to integrals..so I'm sure that doesn't help..
 
  • #4
All constant-acceleration problems can be solved using these two equations:

[tex]x = x_0 + v_0 t + \frac{1}{2} a t^2[/tex]

[tex]v = v_0 + a t[/tex]

All those other equations can be derived from these two. So if you practice how to apply these two equations to a wide variety of problems, you shouldn't have to worry about the other equations. You'll effectively be re-deriving them as necessary.

The one "catch" in using this method is that you sometimes have to solve two equations in two unknowns, so you may need to brush up on that.
 
  • #5
jtbell said:
All constant-acceleration problems can be solved using these two equations:

[tex]x = x_0 + v_0 t + \frac{1}{2} a t^2[/tex]

[tex]v = v_0 + a t[/tex]

All those other equations can be derived from these two. So if you practice how to apply these two equations to a wide variety of problems, you shouldn't have to worry about the other equations. You'll effectively be re-deriving them as necessary.

The one "catch" in using this method is that you sometimes have to solve two equations in two unknowns, so you may need to brush up on that.
On the side note, the equation (2) is just a derivative of equation (1). I.e.

[tex] \frac{dx}{dt} = \frac{d}{dt}\left(x_0 + v_0 t + \frac{1}{2} a t^2\right)[/tex]

[tex]\frac{dx}{dt} = v = v_0 + a t[/tex]

So if you know how to do derivative. You only need one equation :smile:
 
  • #6
kcirick said:
On the side note, the equation (2) is just a derivative of equation (1). I.e.

[tex] \frac{dx}{dt} = \frac{d}{dt}\left(x_0 + v_0 t + \frac{1}{2} a t^2\right)[/tex]

[tex]\frac{dx}{dt} = v = v_0 + a t[/tex]

So if you know how to do derivative. You only need one equation :smile:

I would assume that jtbell is aware of that fact. :smile:
 
  • #7
If you want to go the calculus route, the only equation you ever really need is a = constant! Just integrate "a" twice with respect to t and you get the equations for v and x. :wink:

But take Torec_Scrail's problem as an example:

ball stays in air for 8 seconds being shot straight up, what was its intial velocity? I just took half and basically saw how fast it hit the ground from it's highest point..

I drew a diagram illustrating all the variables, showing which ones are known and unknown, then substituted into the x and v equations and solved them. See the attachment...
 

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1. How do you determine which variables to use in your experiments?

As scientists, we carefully select our variables based on the specific problem we are trying to solve. We consider factors such as the purpose of the experiment, previous research, and the availability of resources. Ultimately, the chosen variables should have a direct impact on the outcome of the experiment.

2. How do you ensure that your formula sheet is comprehensive and accurate?

Before conducting any experiment, we thoroughly research and review existing formulas and theories related to our problem. We also consult with other experts in the field to ensure our formula sheet is comprehensive and accurate. Additionally, we constantly update and revise our formula sheet as new information and discoveries emerge.

3. Can you explain the importance of having only two variables in an experiment?

Having only two variables allows us to isolate and control the factors that may affect the outcome of the experiment. This allows us to accurately determine the relationship between the variables and draw meaningful conclusions. Moreover, it simplifies the data analysis process and makes it easier to identify patterns and trends.

4. How do you handle unexpected results when working with only two variables?

Unforeseen outcomes are a common occurrence in scientific experiments. When dealing with only two variables, we carefully analyze the data to determine the potential reasons for the unexpected results. We may also repeat the experiment multiple times to ensure the accuracy of the results. If the unexpected results persist, we may need to revise our hypothesis and make changes to the experiment.

5. Are there any limitations to using only two variables in an experiment?

While using only two variables can simplify the experiment and data analysis process, it may also limit the scope of the study. Some problems may require more variables to fully understand and solve. Additionally, the real world is often more complex than a controlled experiment, so the results may not always be directly applicable to real-life situations.

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