Find Acceleration: Solving v^2=4s-3

[bondgirl007]

Homework Statement

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

Homework Equations

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?

 

[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.

 

[bondgirl007]

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

 

[HallsofIvy]

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

 

[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?