# Finding acceleration of a skier on parabola

1. Feb 9, 2015

### Raghav Gupta

1. The problem statement, all variables and given/known data
A skier travels with a constant speed of 6 m/s along a parabolic path y= x2/20. Find the acceleration of the skier when he is at (10,5). Neglect the size of skier.

2. Relevant equations
dy/dx is slope of parabola.
In a straight line if body has distance covered=x, then velocity (v) = dx/dt and acceleration = dv/dt
3. The attempt at a solution

Slope is 2x/20= x/10.

2. Feb 9, 2015

### BvU

For a straight line it is indeed very easy: constant speed means zero acceleration !
For a curved trajectory I would think the tangential acceleration at constant speed is zero, so I agree your v2/R is the right answer for this exercise. v is known, so all that remains is to find R at (10,5). We need some equation for that. Any ideas ?

3. Feb 9, 2015

### Raghav Gupta

I saw a unusual formula. From where it has come I don't know.
Can you help on that?
$\frac{1}{R}$ =$\frac{d^2y/dx^2}{[1+(dy/dx)^2]^{3/2}}$

4. Feb 9, 2015

### BvU

I can help on where it comes from: a simple google search brings us here :)

5. Feb 9, 2015

### Raghav Gupta

Thanks, But in the link only formula is given with some description.
Can you provide a derivation of it by some link or by yourself?

6. Feb 9, 2015

### BvU

Yes. Google "radius of curvature equation derivation" and end up here or here. :) I like the second one.

7. Feb 9, 2015

### Raghav Gupta

I also liked the second link.:) Gave the derivation in a nice way. Now the question could be solved by plugging in values in formula .
Only one question that remains is that how according to second link you have provided Curvature or kappa is inverse of radius?

8. Feb 9, 2015

### BvU

Curvature is the better word, I suppose. We associate radius with a circle. They show k = 1/R on page 5.

9. Feb 9, 2015

### ehild

You can solve the problem by yourself, by applying the chain rule.
The equation of the path is given, y=x2/20
You know that the components of the velocity are the time-derivatives of the coordinates. vx=dx/dt, vy=dy/dt. Apply chain rule to find vy in terms of vx: vy=2x/20 dx/dt = x/10 vx.
The magnitude of the velocity is given |v|=6, so vx2+vy2=36. Find vx in terms of x from here.
Once you know vx and vy , apply the chain rule again to derive the components of acceleration. Substitute the given coordinates and find the magnitude.

10. Feb 9, 2015

### Raghav Gupta

Thanks BvU for giving wonderful link for derivation of curvature by differential geometry.

Thanks ehild for showing me another way.
Though it is a lengthy method comparing with curvature formula
but is elegant because I think many do not know the curvature formula by differential geometry.