# Simple math question proportions

#### Maverick56

Lets say you have:

KE = 1/2mv^2 and then you double KE. The velocity increases by a factor of square root 2. Can someone show the algebraic steps to this. Thank you

#### Trayshawn

5+ Year Member
solve for v first. v=sqrt(2KE/m). Replace KE with 2*KE and see what happens. You get an extra factor of sqrt(2). Thats how much the velocity increased.

#### Shjanzey

5+ Year Member
Lets say you have:

KE = 1/2mv^2 and then you double KE. The velocity increases by a factor of square root 2. Can someone show the algebraic steps to this. Thank you
The equation for KE is K = (1/2)*mv^2

So you solve for v

v1 = sqrt(2K/m)

Now double K (kinetic energy) to see what happens

v2 = sqrt((2*2K)/m) --> sqrt(4K/m)

If you compare the two, v2 has increased by sqrt(2).

##### MS-4
Moderator Emeritus
7+ Year Member
I would think about it like this:

You can't increase mass, so you must increase velocity, right? We want the total KE doubled, so:

KE = (1/2)*m*(v)^2

needs to become

KE = 2*(1/2)*m*(v)^2

This means we need to put something into the parentheses that contain "v" that, when squared (and thus pulled out from the parentheses under the exponent), will give us a coefficient of 2. What can we put in as a coefficient of "v" that will be squared to yield 2? The square root of 2 of course.

#### Hemorrage

##### Ambrose
7+ Year Member
I would think about it like this:

You can't increase mass, so you must increase velocity, right? We want the total KE doubled, so:

KE = (1/2)*m*(v)^2

needs to become

KE = 2*(1/2)*m*(v)^2

This means we need to put something into the parentheses that contain "v" that, when squared (and thus pulled out from the parentheses under the exponent), will give us a coefficient of 2. What can we put in as a coefficient of "v" that will be squared to yield 2? The square root of 2 of course.
Very intuitive method

#### MD9

Hopefully this is what you're looking for. If you need any more explanation, please let me know! You can see that instead of doubling the K.E., if we generalize and say we multiply it by some number n, then the result will be that the new velocity, v2, is equal to the initial velocity, v1, multiplied by the square root of n (i.e. just replace "2" in "2K.E." with the variable "n" to get n*K.E. and the result should be v2=sqrt(n)*v1).

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