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Justification for Taking the Positive Square Root in M3

In the kinematics section of Edexcel's M3 module, there is a formula introduced that relates acceleration to velocity when acceleration is a function of displacement. That is:

a=v\frac{dv}{dx}=\frac{d}{dx} (\frac{1}{2}v^2)

.

Often this sort of question may ask you to find

v

in terms of

x

which will involved integrating both sides and then taking square roots.

How do you justify taking the positive square root and disregard the negative one?

Reply 1

davros

Original post by JRichardson12

In the kinematics section of Edexcel's M3 module, there is a formula introduced that relates acceleration to velocity when acceleration is a function of displacement. That is:

a=v\frac{dv}{dx}=\frac{d}{dx} (\frac{1}{2}v^2)

.

Often this sort of question may ask you to find

v

in terms of

x

which will involved integrating both sides and then taking square roots.

How do you justify taking the positive square root and disregard the negative one?

It's probably dictated by the physics of the situation, but to comment properly we'd need to see a specific example!

Reply 2

JRichardson12

Original post by davros

It's probably dictated by the physics of the situation, but to comment properly we'd need to see a specific example!

Sometimes it can be easy. For example the question may say the particle travels on the positive x-axis only. I guess my question is about how much detail generally to give for the ones that are slightly less obvious and what I should be thinking about to justify it to myself.