# Finding the acceleration at any time

#### meeklobraca

1. Homework Statement

The relationship between displacement s and velocity v is given by 1/3s^2 - 4v^2 = 12. Find the acceleration a at any time.

2. Homework Equations

3. The Attempt at a Solution

I know that velocity is found with the derivative of the displacement function, and acceleration is the derivative of the velocity function. Im not sure how to go about it with this type of function.

Up to now the type of functions Ive delt with have been something like s = 2t^2-5t+4 etc

Thank you for your help!

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#### Mark44

Mentor
1. Homework Statement

The relationship between displacement s and velocity v is given by 1/3s^2 - 4v^2 = 12
Is there more to this problem. It doesn't seem to be asking you to do anything.
2. Homework Equations

3. The Attempt at a Solution

I know that velocity is found with the derivative of the displacement function, and acceleration is the derivative of the velocity function.
But velocity and acceleration are not just any old derivatives: they are time derivatives. In particular, v = ds/dt. This might be a hint.
Im not sure how to go about it with this type of function.

Up to now the type of functions Ive delt with have been something like s = 2t^2-5t+4 etc

Thank you for your help!
In anticipation of what the problem might be asking, I'm guessing that it asks you to find s(t) based on the equation you gave. If you use my hint, you'll get an equation that involves s and ds/dt, which maybe you can solve.

#### meeklobraca

Ive editted the question. Im sorry I did leave out what its asking for.

I get that v = ds/dt, and a = dv/dt but as I said the functions ive dealt with are pretty straight forward. I dont know where to start when there is s and v in the same function. Where do I start for this question?

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#### meeklobraca

Upon some investigation and further study, I got a = (1/12)s for this question.

Am I in the ball park with that one?

#### HallsofIvy

Science Advisor
Homework Helper
You are given that 1/3s^2 - 4v^2 = 12. Unfortunately, without parentheses it is not clear whether that first term is (1/3)s^2 or 1/(3s^2)= (1/3)s^(-2). In either case, differentiate with respect to t. Since acceleration is dv/dt, solve for dv/dt, remembering that v= ds/dt.

#### meeklobraca

The first term is (1/3)s^2. I did what you suggested and I got (2/3)s / 8 which = (1/12)s. What do you think?

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