How can I solve for v in this equation?

So your original equation becomesv=\sqrt{ \frac{4l^2}{T^2- \frac{4l^2}{c^2}}}In summary, to isolate the speed of the ship, v, in the given equation, you can multiply both sides by the denominator on the right hand side of the equation and then bring v inside the square root. From there, you can simplify the equation and take the square root to solve for v. In the other example, you can add 1/c^2 to both sides, take the inverse, and then take the square root to isolate v.
  • #1
Pengwuino
Gold Member
5,123
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I have this equation here and I don't see how I can isolate v, the speed of the ship...

[tex]\frac{{2(4ly)(9.5x10^{15} \frac{m}{{ly}}}}{v} = \frac{{16y}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}[/tex]

y = years
ly = light years

The book doesn't tell you how it got v=0.447c, it just jumped to the conclusion after presenting the equation. It seems like no matter how I try the equation, the v-squared will always cancel and i'll lose my velocity equation. Am I not doing this correctly or is there a more complex method of solving for v?
 
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  • #2
I will do some cleaning up:

[tex] a = 2(4ly)(9.5x10^{15}) [/tex]

[tex] b=16y [/tex]

so we write it much cleaner as:

[tex] \frac{a}{v} = \frac{b}{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} } [/tex]

Now I just multiply by the denominator of the RHS and get:

[tex] \frac{a}{v} [\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }] = b [/tex]

Now I just bring the v inside the square root:

[tex] a[\sqrt {\frac{1}{v^2} - \frac{{1}}{{c^2 }}}}] = b [/tex]

No v's cancel out.

You can simply it to get,

[tex] ( \frac{1}{v^2} - \frac{1}{c^2}) = \frac{b^2}{a^2} [/tex]


[tex] \frac{1}{v^2} = \frac{b^2}{a^2} + \frac{1}{c^2} [/tex]

invert both sides and take square root:

[tex] v = +/- \sqrt{(\frac{b^2}{a^2} + \frac{1}{c^2})^{-1}} [/tex]
 
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  • #3
Just from eying the equation, wouldn't squaring both sides, then multiplying both sides by v²(1-(v/c)²) give you a quadratic equation in v²? the uniqueness of the v would then be due to the fact that one of the root is negative, so not a physically acceptable solution.
 
  • #4
Ok I have another problem where I can't seem to isolate v correctly... A rocket will go to a galaxy 20 light years away and it can only take 40 years round trip according to the rocket passengers. I figured this is how i could write it.

[tex]\begin{array}{l}
T' = \frac{{2L}}{v} = \frac{{T_0 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }} \\
\sqrt {1 - \frac{{v^2 }}{{c^2 }}} *2L = T_0 *v \\
\sqrt {4l^2 - \frac{{4l^2 v^2 }}{{c^2 }}} = T_0 v \\
4l^2 - \frac{{4l^2 v^2 }}{{c^2 }} = T^2 v^2 \\
4l^2 (1 - \frac{{v^2 }}{{c^2 }}) = T^2 v^2 \\
\frac{1}{{v^2 }} - \frac{1}{{c^2 }} = \frac{{T_0 ^2 }}{{4l^2 }} \\
\end{array}[/tex]

Now I am not sure how to isolate v
 
  • #5
Add 1/c^2 to both sides, take the inverse, then take the square root, exactly as before.
 
  • #6
How do i do the inverse of two added quantities though?
 
  • #7
make them a common fraction and do the addition. Then take the inverse of that new fraction you get.
 
  • #8
How would i do that... man i can't do basic math!
 
  • #9
Do you do this?

[tex]v^2 = \frac{{4l^2 + c^2 }}{{T_0 }}
\][/tex]
 
  • #10
In your latter example you would get:

[tex] \frac{T^2}{4L^2} + \frac{1}{c^2} [/tex]

[tex] \frac{c^2T^2}{4L^2c^2}+ \frac{4L^2}{4L^2c^2} [/tex]

[tex] \frac{c^2T^2+4L^2}{4L^2c^2} [/tex]

The inverse would be:

[tex] \frac{4L^2c^2}{c^2T^2+4L^2} [/tex]
 
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  • #11
Dang that's insane... that's from like 5th grade :( and i forgot!
 
  • #12
Ok and since I know the space ships speed to be 0.894c... does that mean the time measured from the stationary observers on Earth is ~2.236*40 years using gamma*T proper?
 
  • #13
Sorry, I am not a physics major. Thats as much help as I can provide.
 
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  • #14
Pengwuino said:
Ok I have another problem where I can't seem to isolate v correctly... A rocket will go to a galaxy 20 light years away and it can only take 40 years round trip according to the rocket passengers. I figured this is how i could write it.

[tex]\begin{array}{l}
T' = \frac{{2L}}{v} = \frac{{T_0 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }} \\
\sqrt {1 - \frac{{v^2 }}{{c^2 }}} *2L = T_0 *v \\
\sqrt {4l^2 - \frac{{4l^2 v^2 }}{{c^2 }}} = T_0 v \\
4l^2 - \frac{{4l^2 v^2 }}{{c^2 }} = T^2 v^2 \end{array}[/tex]
You are great up to this point but remember you want to solve for v- you want to isolate it, not mix it with other things.
[tex]4l^2= T^2v^2- \frac{4l^2}{c^2}v^2= (T^2- \frac{4l^2}{c^2})v^2[/tex]
[tex]v^2= \frac{4l^2}{T^2- \frac{4l^2}{c^2}}[/tex]
 

FAQ: How can I solve for v in this equation?

1. What is the equation for calculating distance traveled at the speed of light?

The equation for calculating distance traveled at the speed of light is: distance = speed of light x time. This equation is based on the principle that the speed of light, denoted by the letter c, is a constant and is equal to approximately 299,792,458 meters per second.

2. Can anything travel faster than the speed of light?

According to the current understanding of physics, it is impossible for anything to travel faster than the speed of light. This is because the speed of light is a fundamental constant in the universe and as an object approaches the speed of light, it would require infinite energy to reach it, making it impossible to exceed.

3. How does the concept of time dilation affect distance traveled at the speed of light?

Time dilation refers to the phenomenon where time appears to slow down for an object as it approaches the speed of light. This means that from the perspective of an observer, an object traveling at the speed of light would appear to take longer to cover a certain distance compared to an object traveling at a slower speed. Therefore, from the perspective of the object traveling at the speed of light, the distance traveled may not seem as great as it does to an observer.

4. Is it possible to travel at the speed of light?

While it is currently not possible for any object with mass to travel at the speed of light, it is theoretically possible for particles with no mass, such as photons, to travel at this speed. However, this is only possible in a vacuum and any object with mass would require infinite energy to reach the speed of light, making it impossible to achieve in practice.

5. How does the theory of relativity impact distance traveled at the speed of light?

The theory of relativity, proposed by Albert Einstein, states that the laws of physics are the same for all observers in uniform motion. This theory has a significant impact on the concept of distance traveled at the speed of light. It explains how an observer's perception of distance and time can differ from that of an object traveling at the speed of light, and how the speed of light is a fundamental constant in the universe.

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