How fast could the person have to travel?

[hage567]

i know this is stupid question...i just came across...do break the square root u have to square it then its (v2/c2) ^2..isnt it

If you have

\sqrt{1 - v^2/c^2}

and square it, you get 1 - v^2/c^2 NOT (1 - v^2/c^2)^2

 

[jenita]

lol...enways is my answer right...can u check it...if it is then thank yooooooouuuuuuuuuuu so much...

 

[hage567]

The 2.4 x 10^8 m/s? No, that's not what I get.

 

[jenita]

omg this is killing me...let me see what i did wrong

 

[jenita]

ok let me check i did 48/60 and i got .8 and squared it after that i squared C and multiplied with .64 and got that answer and i did what u told me to do it...

 

[hage567]

\sqrt{1 - v^2/c^2} = \frac{M_o}{M}

See the 1 under the square root on the left side? What happened to that?

 

[jenita]

you subtract that and still get the same answer

 

[jenita]

hold on a minute let me do it

 

[jenita]

i give up...i have 134164078.6 now

 

[hage567]

you subtract that and still get the same answer

You do?

Until you can show me a picture or write out your equation in latex I don't think I am going to be see what you are doing wrong.

I think you are just getting caught up in some small math error.

 

[hage567]

v = \sqrt{1-(\frac{M_o}{M})^2} c

This is what you should get after arranging your equation. Please make sure you understand how to get this.

Note: the c is not under the square root.

 

[jenita]

let me draw and sent it to you

 

[jenita]

hold on how did u get (mo/m) square

 

[hage567]

I don't have much time to do it now, it's getting late. If you send it I can try to check it in the morning.

 

[hage567]

hold on how did u get (mo/m) square

Look at post #25. To get rid of the square root, you must square both sides of that equation.

Also see what I said in post #31.

 

[jenita]

do u substitute mo/m in v2

 

[jenita]

did u get 297993288.5 the answer...if this is wrong i am so giving up

 

[hage567]

do u substitute mo/m in v2

NO. Why would you do that?? You want to solve for v. Why would you want to eliminate it from the equation? You are not supposed to be solving for c, if that's what you're up to.

did u get 297993288.5 the answer...if this is wrong i am so giving up

No that's not the right answer.

I gave you the equation for v. Put the numbers in and see what you get. But you must ask for help (perhaps from your teacher) if you can't see how to work out that answer.

Anyway, I must go. Let me know what you come up with and I will check it tomorrow.

Good night.

 

[jenita]

good night
and thanks for the help

 

[Aikenfan]

I got:

V = Square root (1-(48/60)^2 (c)
=60,000,000

Is that correct?

 

[jenita]

i got 1.8*10^8

 

[hage567]

I got:

V = Square root (1-(48/60)^2 (c)
=60,000,000

Is that correct?

No, that's not right. You need to both square (Mo/M), and then take the square root of
(1-(Mo/M)^2). It doesn't seem like you did that correctly. At least that's what it looks like to me.

 

[hage567]

i got 1.8*10^8

This is the answer I get.

 

[jenita]

thanks though...we got the correct answer its .6c ...i know what i did wrong... i already turned in my paper...thanks a looot...u were so helpfull...

 

[hage567]

You're welcome. I'm glad you figured out what was going wrong!

 

[jalalmalo]

M0 is 48 and M is 60 the speed of light is 3x10 to the power of 8, insert everything in the equation and solve for v and you'll have the answer lol.

 

[regor60]

wow...

 

[malawi_glenn]

M0 is 48 and M is 60 the speed of light is 3x10 to the power of 8, insert everything in the equation and solve for v and you'll have the answer lol.

this thread is over 2 years old...

 

[jalalmalo]

:P hehe my bad