Table 2-4 (p. 34) in your text gives a set of equations that apply to
objects undergoing uniform acceleration. These can be used to solve
one of the three major problem types that we'll encounter in mechanics. We'll call this type a dvat problem, where the acronym stands for
position-velocity-acceleration-time. The equations are listed below
with descriptions.
1 |
v = vo + at |
This is simply an algebraic rearrangement of the definition of
acceleration: a = Δv/Δt.
Galileo defined acceleration in this way, because he found that objects
rolling down inclines increased in speed by equal amounts in equal
time intervals. By defining acceleration as a = Δv/Δt,
he could say that these objects had constant acceleration. While
he could have selected some other formula such as a = Δv/Δx,
that wouldn't have been useful in describing the motion of real objects. |
2 |
vav = ½(vo + v) |
This is called the Merton Theorem and was developed at Merton College
of Oxford in the 13th century. Basically, it amounts to saying
that when a quantity increases at a uniform rate, the average value
of the quantity is equal to the middle value. For example, suppose
you had the series of numbers 1,2,3,4,5. Add them up and divide by
5 to get 15/5 = 3, the average. But this is also the middle
number in the sequence. Note that the numbers in the sequence
increase by equal amounts. Try applying the Merton Theorem to
the sequence 1,3,4,5, and you'll see it doesn't work for that non-uniform
sequence. |
3 |
x = xo + ½(vo +
v)t |
This is a combination of the second equation and the definition
of average velocity, vav = Δx/Δt. |
4 |
x = xo + vot + ½at² |
These equations are derived algebraically from combinations
of the first three equations. Note that the sixth equation is
one that isn't in the book. However, it can be useful in solving
some problems. |
5 |
v² = vo² + 2a(x - xo) |
6 |
x = xo + vt - ½at² |
-
The quantities subscripted with o indicate initial
values, while those that don't have subscripts represent final
values. The initial value of the time is assumed to be
zero.
-
Times are always positive or zero.
-
The equations apply only to objects with constant or zero acceleration.
-
The equations apply to objects moving in a straight line.
- The quantities x, v, a, and their
initial values may be positive or negative. The sign is
determined by the direction selected for the positive x-axis. Suppose the positive x-axis is set up to be pointing
to the right. In this case, an object moving to the right has
a positive velocity, and an object moving to the left has a negative
velocity. Consider the following four possible situations.
Description of
the motion in words |
Sign of v |
Direction of v |
Sign of a |
Direction of a |
Description of
the motion in mathematical terms |
Object is moving to the right and the magnitude
of the velocity is increasing
(object is speeding up to the right) |
+ |
---> |
+ |
---> |
The velocity is becoming more positive, so
Δv in Δv/Δt is
positive. |
Object is moving to the right and the magnitude
of the velocity is decreasing
(object is slowing as it moves to the right) |
+ |
---> |
- |
<--- |
The velocity is becoming less positive, so
Δv in Δv/Δt is
negative. |
Object is moving to the left and the magnitude
of the velocity is increasing
(object is speeding up to the left) |
- |
<--- |
- |
<--- |
The velocity is becoming more negative, so
Δv in Δv/Δt is
negative. |
Object is moving to the left and the magnitude
of the velocity is decreasing
(object is slowing to the left) |
- |
<--- |
+ |
---> |
The velocity is becoming less negative, so
Δv in Δv/Δt is
positive. |
-
As is frequently the case in physics, it's not necessarily
useful to memorize the above statements. If you understand
where the relationships come from, then you can produce them. That's why we provide the descriptions in mathematical terms
above to help you understand how the sign of the acceleration
is determined.
-
Something important to realize is that a negative acceleration
needn't mean that an object is going slower and slower. The
negative sign is determined by the choice of positive x,
which in turn determines the direction of positive v.
|
With the above in mind, we'll look at a simple strategy
to solve dvat problems. We'll describe the strategy first and
then apply it to a specific situation.
Step 1. After
reading the problem, draw a diagram. On the diagram, indicate the
origin, the direction you select for +x, and the corresponding
directions for initial velocity and acceleration. Label any other
relevant quantities.
Step
2. List all the given information. Identify the
givens with the same symbols that are used in the dvat equations,
namely, x, xo, v, vo,
vav, and t. Given the direction you
selected for +x, make sure all the given information has
the correct sign.
Step
3. State the unknown that you're supposed to find.
Step
4. Look at the list of dvat equations and select
one for which all quantities are known except for the unknown that
you're solving for.
Step
5. Algebraically solve the dvat equation you
selected for the unknown. That means to solve in symbolic form without
numbers.
Step
6. Substitute the given values with units. Do the arithmetic
to arrive at the final answer.
Step
7. Apply sign, units, and sensibility checks. |
Now let's apply the strategy to a problem. Click
here to go on.
|