Changing factor of centripetal acceleration

[dandy9]

Homework Statement

A particle moves in a circular path of radius r with speed v. It then increases its speed to 19v while traveling along the same circular path. By what factor has the centripetal acceleration of the particle changed?

Homework Equations

ac=v^2/r

The Attempt at a Solution

Tried solving for v in terms of a and r to no avail.

 

[Doc Al]

What's the centripetal acceleration when the speed is v? When the speed is 19v? Compare.

 

[PhanthomJay]

If the initial centripetal acceleration is v^2/r, then try substituting 'v' with '19v', to get the new centripetal acceleration, then compare the 2 values.

 

[AndyNewton]

Tried solving for v in terms of a and r to no avail.

This question is so ridiculously simple that it creates a thinking-trap! It is tempting to think "It can't be that easy... must do some difficult equation solving ... etc."
It is good to know that things are not always complicated. Sometimes the answer is right in front of you

The question asks about centripetal acceleration.
You already have a "relevant equation" giving an expression for centripetal acceleration.

 

[AndyNewton]

If you can see the obvious answer now, that's fine.

If you still can't see the answer I will point out one little criticism of the original question, or at least the way it is entered in message #1. This is a possible source of confusion:

The question contains a term "v" and another term "19v". The relevant equation as quoted above also contains a term "v".
It would be better if the relevant equation was written as acc=(velocity)^2 / r

Then it becomes more natural to recognize that you can insert any value into the (velocity) part of the equation.
The (velocity) term in the equation is initially equal to "v" and then it changes to be equal to "19v".

 

[dandy9]

Wow, a lot of replies.
Thanks a lot everyone. I understand it now! (And I was able to tell my friends who were also stumped on this question.)
Thanks again!