Solving for a and t Given s, vi, and vf

  • Thread starter JasonRox
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In summary, the conversation discusses using variables and derivatives to solve a differential equation with initial conditions. The conversation also touches on the validity of using different methods to solve the problem. The participants agree that the method shown is correct but caution that it is important to clearly state the problem being solved.
  • #1
JasonRox
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Here is the info.

s = 150m
a = ?
t = ?
vi = 0
vf = 27

Here is what I did:

[tex]27 = at[/tex]
[tex]27/t = a[/tex]

Put a into another unsolved equation.

[tex]150 = 1/2(27/t)t^2[/tex]
[tex]11.1 = t[/tex]

Put the known variable into the first equation.

[tex]27 = a(11.1)[/tex]
[tex]2.43 = a[/tex]

All I did was take variables, and insert it into it's derivative, or antiderivative. I'm getting different answers using other methods. Can someone explain if this is incorrect?
 
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  • #2
Your algebra seems correct, how ever your arithemtic is suspect.

I get t=11.1s
 
  • #3
Yeah, you're right.

Human error. :biggrin:

So, even though you are matching up numbers of different graphs, it still is valid?
 
  • #4
If I understand your question correctly, the answer is yes.

Actually you are solving the Differential Equation:

[tex]\frac {d^2s} {dt^2} =a [/tex]
with initial conditions

[tex]\frac {ds(T)} {dt}= 27 \frac m s [/tex]

[tex] \frac {ds(0)} {dt}= 0 \frac m s [/tex]

[tex]s(T) =150m[/tex]

[tex]s(0)=0[/tex]

T is the time you need to find.

This can be solved by integration, with each integration constant evaluated using the initial conditions. So your "velocity graph" is the derivative of the "position graph" and are thus simply showing different aspects of the same physical situation.
 
  • #5
The difficulty in telling you if this method works for solving the problem is that you never actually told us what the problem is! You said you did this and that but what is it you are trying to solve?
 
  • #6
I gave you the info. I had, and showed my steps.

I wasn't really concerned on finding the answer. I was just curious if using certain variables into other equations to find the answer. I know you can do this, but I wasn't quite sure if you can do it with its own derivative.

I guess you can. Thanks for the advice.
 

1. How do I solve for a and t given s, vi, and vf?

To solve for a and t, you can use the equation s = vit + 1/2at^2, where s is the displacement, vi is the initial velocity, and vf is the final velocity. Rearrange the equation to solve for either a or t depending on which one is unknown, then plug in the values for s, vi, and vf to find the solution.

2. What units should I use for the given variables?

The units for s should be in meters, vi and vf should be in meters per second, and a should be in meters per second squared. It is important to use consistent units throughout the equation to ensure accurate results.

3. Can I solve for a and t if I only have two of the given variables?

No, you need all three variables (s, vi, and vf) to solve for both a and t. If you are missing one of the variables, you will not be able to find a unique solution.

4. What if I am given multiple values for s, vi, or vf?

If you are given multiple values for any of the variables, you can use the average value in the equation. For example, if you are given two values for s, you can find the average and use that value in the equation to solve for a and t.

5. Can this equation be used for any type of motion?

This equation can be used for motion with constant acceleration, such as an object falling due to gravity. It may not be applicable for other types of motion, such as circular motion or motion with changing acceleration.

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