# Motion Problems

1. Oct 23, 2007

### bondgirl007

1. The problem statement, all variables and given/known data

If v$$^{2}$$=4s-3, find the acceleration when t=1s.

2. Relevant equations

3. The attempt at a solution

I know that acceleration is the derivative of velocity, which is the derivative of distance but how do I get the derivative with both v and s in the equation?

2. Oct 23, 2007

### Dick

v is velocity, s is distance or time??? There's a problem with v^2=4s-3. The units on both sides don't appear to match.

3. Oct 23, 2007

### bondgirl007

S is distance but I'm not sure how to solve this problem.

4. Oct 23, 2007

### HallsofIvy

Staff Emeritus
Since, as you say, "acceleration is the derivative of velocity" (with respect to time), you find the acceleration by differentiating wih respect to time.

Differentiating both sides of v2= 4s- 3 with respect to time: using the chain rule, of course: 2v v'= 4s'. Now v'= a, the acceleration, and s'= v, so that equation is
2va= 4v and a= 2, a constant!

Dick, if you assume the "4" and "3" given are dimensonless then, yes, the right side has units of "distance2/time2" while the right side has two parts: one with units of "distance/time" and the other dimensionless- so they can't even be combined. However, if we assume that "4" is really "4 m/s2" and the "-3" is really "-3 m2/s2",then we are alright

5. Oct 23, 2007

### bondgirl007

Thank you so much!

I have another question similar to this and was wondering if I did it right. Here's the question and my attempt:

s$$^{2}$$ - 6v$$^{2}$$ = 10. Find the acceleration in terms of the distance.

2s (ds/dt) - 12v dv/dt = 0
2s (ds/dt) - 12(ds/dt)(a) = 0
2s (ds/dt) = 12(ds/dt)(a)

a = s/12.

Am I right?