Physics - Acceleration
Date: 09/13/2001 at 22:42:47 From: Nathan Shehorn Subject: Physics--acceleration What equation do you use to find the time element when you have the change in velocity and the displacement?
Date: 09/14/2001 at 09:54:25 From: Doctor Rick Subject: Re: Physics--acceleration Hi, Nathan. That's not sufficient information, unless when you say "the change in velocity" you mean both the initial and final velocities. I will assume you mean this. The general equations of uniformly accelerated motion in one dimension are x = x_0 + v_0*t + at^2/2 v = v_0 + at The displacement from time 0 to t (which I'll call d) is x - x_0, so we can write d = v_0*t + at^2/2 I'll call the velocity at time t the final velocity, v_f. You want to find t, knowing d, v_0 and v_f, but not a. We can solve the velocity equation for a: a = (v_f - v_0)/t Substitute in the displacement equation: d = v_0*t + (v_f - v_0)t/2 = (v_0 + v_f)t/2 This is easy to solve for t: t = 2d/(v_0 + v_f) Looking back at the previous equation, we see that the displacement is the time interval times (v_0 + v_f)/2, and this quantity is the mean (or average) of the initial and final velocities. Thus our final formula can be stated this way: The time interval is the distance divided by the mean of the initial and final velocities. Now recall the rate equation, which relates displacement, time, and velocity in the case of motion at constant velocity. v = d/t Looking at this another way, we *define* the "mean velocity" (even if the velocity changes) as the total displacement divided by the time interval. _ v = d/t For uniformly accelerated motion, we find from the equation for d above that _ v = (v_0 + v)/2 The mean velocity over the whole time interval is the mean of the initial and final velocities. That's easy to remember. It isn't true in general, though; only for motion with constant acceleration (including zero acceleration, of course). - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.