Problem 12-10 is escaping me for some reason.

There is a particle moving with initial velocity of 20 m/s. The acceleration of the particle is a = -2*v m/s^2. Find the velocity as a function of position and the displacement of the particle when it stops.

I've tried using the relation a ds = v dv, as well as the other differential relations, but I am coming up empty-handed. Can anyone help?

1/13/2008 10:56:25 PM

Might be too late but I'll post anyway.

Velocity is the derivative of the position function 'S' and acceleration is the derivative of the velocity function 'V'. You're given the acceleration function 'A' which is constant at -2 m/s^2. You're also given the initial value for velocity (where V(0) = 20 m/s).

To get the velocity you take the integral of the acceleration function and get V(x) = -2 * x + C.
You solve for C of course by plugging in x = 0, so you get V(0) = C = 20m/s, so C = 20 m/s.

You take the integral of V(x) = - 2 * x + 20 to get the position function S and end up getting:

S(x) = - x ^ 2 + 20 * x

To find the displacement of the particle when it stops, solve V(x) = 0 for x where x > 0 and plug it into S(x).

1/14/2008 12:47:47 AM

I was in class on Friday, but don't know if he assigned hw. Does anyone else know (actually, I'm in the first class).

2/10/2008 7:45:07 PM

pay attention!!!!

2/15/2008 12:33:59 AM

^^^Isn't the acceleration -2*v, not -2, therefore A isn't constant?

[Edited on February 15, 2008 at 7:59 AM. Reason : ^]

2/15/2008 7:58:42 AM