Converting the Speed of Light into Micrometres/Year for Radical Equation Problem

  • Thread starter Stratosphere
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In summary, the conversation discusses solving the equation 86300=\frac{86400}{1/\sqrt{1-(v/299000000)^2}} for the variable v, which involves dividing both sides by 86400 and then multiplying both sides by the square root of 1 - (v/299000000)^2. It also suggests using units of measurement such as c or micrometres/year instead of metres/second to avoid astronomically high numbers in the solution.
  • #1
Stratosphere
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Homework Statement


86300=[tex]\frac{86400}{1/\sqrt{1-(v/299000000){2}}}[/tex]


Homework Equations





The Attempt at a Solution


When I square each side I get an astronomicly high number.
 
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  • #2
hi stratosphere

have you tried dividing both sides by 86400 first?
 
  • #3
Or, better, dividing both sides by 86300 and then multiplying both sides by that square root.

And, is that
[tex]\sqrt{1- (v/299000000)2}[/tex]
supposed to be
[tex]\sqrt{1- (v/299000000)^2}[/tex]?
 
  • #4
HallsofIvy said:
And, is that
[tex]\sqrt{1- (v/299000000)2}[/tex]
supposed to be
[tex]\sqrt{1- (v/299000000)^2}[/tex]?

Yah I meant to make it an exponent.
 
  • #5
If you're attempting to use the relativity equations (I'm guessing to find length contraction?), rather than using metres/second for the speed of light and resulting in astronomically high numbers, or better yet, converting the speed of light into micrometres/year; why not just leave it as c, and after solving for c, substitute whatever unit of measurement you want in place for it.
 

1. What is a radical equation problem?

A radical equation problem is an algebraic equation that contains a variable within a radical symbol, such as a square root or cube root. These types of equations often involve finding the value of the variable that makes the equation true.

2. How do you solve a radical equation problem?

To solve a radical equation problem, you must isolate the radical on one side of the equation and then square both sides to eliminate the radical symbol. This process may need to be repeated multiple times if there are multiple radical symbols in the equation.

3. Can radical equations have extraneous solutions?

Yes, radical equations can have extraneous solutions, which are values that appear to be solutions to the equation but do not actually satisfy the original equation. This can happen when squaring both sides of the equation introduces additional solutions.

4. Are there any restrictions when solving radical equations?

Yes, when solving radical equations, it is important to check for any restrictions on the variable. This could include the variable having to be non-negative or not equal to zero in order for the equation to be valid.

5. How do radical equations relate to real-life situations?

Radical equations can be used to model real-life situations, such as calculating the time it takes for a substance to decay or the distance an object travels in a given amount of time. These equations can also be used in engineering and physics to solve for unknown values in various formulas and equations.

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