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Physics - AccelerationDate: 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/
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