| Week of | Aug 23
| Sep 6
| Oct 2
| Nov 6
| Phy 205 page |
| Aug 28 | Sep 11 | Oct 9 | Nov 13 | ||
| Sep 18 | Oct 16 | Nov 20 | Notes for Fall, 1999 | ||
| Sep 25 | Oct 23 | Nov 27 | |||
| Oct 30 |
11 Sep
The two key equations in chapter two are
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To see how to use them, start with an example and walk through the definitions. Take
A, B, and C are three constants, and I picked particular numerical
values in class. Apply the definition of velocity to this and you get
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When you expand and simplify this, you get
Finally, take the limit as Dt approaches zero and the middle term goes away, leaving the function
A similar set of calculations for the acceleration, gives
When you divide this by Dt, you get 2A (even before taking a limit). In this special case, the function ax(t) happens to be a constant.
In most cases you don't know the position of an object as a function of time; that may be what you are trying to find out! The problem is usually the reverse of what I've just done. Instead of finding the derivative of a function, it is often necessary to find the anti-derivative.
The anti-derivative of a function is the function whose derivative is what you started with.
The simplest common example of this is gravitational free
fall. If the only force on a object is from gravity, then the
acceleration (near the Earth's surface) is constant. I worked out the
example of throwing a baseball straight up at an initial speed of say
50 mph = v0. The experimental observation about gravity
near the Earth is that the acceleration is constant. Choose the x-axis
to be positive upward, then the equation to solve is
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I worked backwards from this to find the velocity and the position as functions of time. There are two arbitrary constants that show up in this calculation. These are determined from the initial conditions stated in the problem. I threw the ball up with a given initial speed and I will select my coordinate system so that the origin is where I threw it from.
The anti-derivative of the equation dvx/dt=-g is
where C is some unknown constant. The next step is to take another anti-deriviative, because the preceeding line is dx/dt:
where D is still another unknown constant.
To evaluate these two unknowns requires two equations. These come from the unused part of the information in the statement of the problem.
I have some freedom here. I chose the coordinate system so that the origin is at the position from which I throw the ball. The two equations are now
These two equations determine C = v0 and D = 0.
Put the equations together with the initial conditions and you have
The future motion of the ball is now known, at least until it hits the ground.
13 Sep Discuss in some detail a homework problem that involves relative velocities. The main difficulty is in picturing what is happening. I drew some pictures that may help. Also, it helps to view the problem from the perspective of the person who is moving.
Recapitulate the problem from Monday, cutting through to the essential
steps. Then do another example the same way: From the old exam,
the airplane takes off, reaching its takeoff speed of v
15 Sep
Do in detail several example problems. One from an old exam, two more
from today's homework.
Solve all these problems the same way.
Start with a sketch and what is given. Use the basic equations
Whether the acceleration is a constant or not, start here and find the
anti-derivatives. Use the initial conditions to determine what the
constants are, and you may get into some messy algebra, but the
concepts are always the same.
1. what is given
2. a crude sketch
3. pick a coordinate system
4. state all the given data in terms of the coordinate system; give
names to the unknowns
5. use the basic definitions of acceleration and velocity, taking
anti-derivatives
6. evaluate the constants from the initial data
7. use the rest of the information given to find equations for
the other two unknowns
8. try some numbers to see if the result is plausible.
ax = dvx / dt