Lecture 2

Sign of vector components, general rule : the value of a vector component is given by the coordinate of the head's projection minus the coordinate of the tail projection.

From this, it follows immediately that a component will be positive if it "points" in the same drection as the axis, negative if it "points" in the opposite direction.

It is obvious that the sign of the components will depend upon the choice of orientation for the reference axes. The normal choice is the x-axis pointing to the right and the y-axis pointing up, but nobody can prevent me from making a different choice (in problems dealing with gravity and falling bodies, it is not uncommon to have the vertical axis pointing downwards).

once you make a choice for a given problem, stick with it throughout all the steps, and evaluate consistently the signs of the various components

Motion in one dimension

We start now the study of MECHANICS, i.e. the branch of Physics that deals with the motion of objects. Mechanics is traditionally didvided into :

• KINEMATICS : "how" things move
• DYNAMICS : "why" things move

We start with a few, rather trivial, definitions :

• Displacement
  if an object moves from a point A to a point B, we call displacement a vector pointing from A to B, with a magnitude given by the distance between A and B.
  
  Do not confuse Displacement with Distance Covered : if I drive from here to Washington and then back, my final displacement is zero, even though I travelled 200 miles...
• Speed
  if an object covers a certain distance d in a time t, we call (average) speed the ratio of the distance to the time needed to cover that distance
  average speed = distance/time = [LT^-1]
  By definition, speed is always a positive quantity, and it is obviously a scalar
• Velocity
  The concept of speed carries no information on the direction of the motion. To convey this information we introduce the concept of (average) velocity, which is defined as the ratio between the displacement vector and the time required to attain the given displacement. In the most general case, if an object is at location $\vec{A}$ at the time t = t₁ and at location $\vec{B}$ at the time t = t₂, its (average) velocity $\vec{v}$ is given by :
  $\vec{v} = (\vec{B-A})/(t_{2}-t{_1})$ = Displacement/(Time to Displace)

These definitions are quite general, and apply to every possible situation. For a while, we will study motions taking place only along a given direction, let's say along the x-axis. For simplicity, in this case we will not need to specify explicitely when we are dealing with vector quantities (the direction is always the same), but let us not forget that quantities like velocity and displacement can be positive or negative.

If our object is at position x₀ at the time t₀ at at position x at time t, the above expression reduces to :

$v = (x-x_{0})/(t-t_{0}) = \Delta x/\Delta t$

Do remember that this represents the average velocity in the time interval $\Delta t$ (or the actual velocity, if the object is traveling at a constant, non changing rate).

In general, objects will not move at constant velocity, and we might like to know their actual velocity at any given instant. Given that $v = \Delta x/\Delta t$, the smaller we make the interval $\Delta t$ during which we measure the displacement, the closer we get to measure the instantaneous velocity. Practically, it is impossible to measure the real instantaneous velocity, but mathematically it is meaningful. The standard definition is

$v_{inst} = lim_{\Delta t\rightarrow 0}~\Delta x/\Delta t$

If you have never dealt with Calculus, you might wonder about the meaning of this expression (what does it mean to divide 0 by 0 ?), but rest assured that the operation is meaningful and it gives a reasonable result (we will see its meaning in a while). But we can at least state the following : if an object is moving with constant velocity, then

velocity = average velocity = instantaneous velocity

When dealing with average (or constant) velocities, we can rewrite the expression above as

x(t) = x₀ + vt

where we have assumed that x₀ is the position at t=0. This expression summarizes everything that we can say about motion with constant velocity (or for which only the average velocity is known), and it allows us to solve any possible problem dealing with such a situation. We can thing of the expression as containing 4 variables, which means that, known any 3 of them, we can uniquely find the 4^th one.

It is very useful to think of the geometrical interpretation of the expression : it represents a straight line in the x-t plane. And the slope of the line, geometrically given by $\Delta x/\Delta t$ is nothing but the velocity !!!

Problems dealing with a single object moving with constant velocity are fairly straightforward. More interesting is the case when we have two objects, each one moving with its own (constant) velocity, and we are asked to find out where (or when) they meet. To solve these problems, it is useful to thing of the meaning of the term "to meet" : it means to be at the same place at the same time. The solution to the problem is then to find the "t" for which the two objects have the same "x". Specifically, the motion of object 1 is given by x(t) = x₀₁ + v₁t, and object 2 by x(t) = x₀₂ + v₂t. If, at any time, they are at the same "x", we must have

x(t) = x₀₁ + v₁t = x₀₂ + v₂t
v₁t - v₂t = x₀₂-x₀₁
t = (x₀₂-x₀₁)/(v₁-v₂

which gives the "meeting time". Substitution in either of the x(t) expression will give me the meeting place... Notice that the expression we have found for t will always have a unique solution, as long as $v_{1}\neq v_{2}$ .... What is the graphical interpretation of the problem ? It is that of two straight lines in the plane, that will always intersect in one point, unless they are parallel...

Motion with constant Acceleration

Now that we know how to handle constant velocities, we can attack the case of variable velocity. Whenever there is a change in velocity, we will say that there is an acceleration.

Let us suppose that an object has a velocity v₀ at time t₀ and v at time t. We will say that its (average) acceleration was

$a = (v-v_{0})/(t-t_{0}) = \Delta v/\Delta t$

(Note that, as we are still working in one dimension, I am not using the vector nomenclature, but let us not be confused : in general acceleration, as given by the difference between two vectors divided by a scalar, is itself a vector)

Acceleration expresses the change of velocity in a given time, therefore it will be measured (in the SI system) in meters per second .... per second = (m/s)/s = m/s², or any other equivalent unit. Dimensionally,

[acceleration] = [L/T²] = [LT^-2].

Note : in common language, one talks about accelerations and decelerations, and one might even be inclined to think that a deceleration is a "negative acceleration". This can be misleading, and it is better to think in the following terms : we have a real acceleration (i.e. the velocity is increasing) when acceleration and velocity point in the same direction (i.e. they have the same sign). Velocity will decrease (i.e. we are "decelerating") when velocity and acceleration have opposite signs

In analogy to what was done earlier, we can think in terms of average acceleration (if we only know the initial and final conditions) or, when the acceleration is continuously varying, of instantaneous acceleration, which, will be defined as previously :

$a_{inst} = lim_{\Delta t\rightarrow 0}~\Delta v/\Delta t$

Throughout the course, we will almost always limit ourselves to the case of constant (or average) accelerations. For this case we have immediately :

v(t) = v₀ + at

whwre we have assumed that, at t=0, the velocity is is v₀. Notice the formal identity between the expressions relating time and displacement (for the case of constant v) and time and velocity (for the case of constant a). The procedures we have learnt for the former situation can be transported to the latter.

But this is not enough, to describe the motion fully we must be able to say something about displacement. Here is how we can do it : to start, we know how to find the displacement if we know the average velocity :

x = x₀ + v_avt

But, in our case of constant acceleration, what is the average velocity? let us say that we start at t=0 with a velocity v₀ and at time t the velocity is v. We have just seen that v(t) = v₀ + at. The average velocity will then be the midway value between initial and final velocity:

v_av = (v_final + v_initial)/2 = (v₀ +at + v₀)/2 = v₀ + at/2

Substitution into the expression for x gives :

x(t) = x₀ + v₀t + 1/2 at²

This expression, together with v(t) = v₀ + at, is all that you need to fully describe the rectilinear motion with constant acceleration, and allow you to solve any possible problem.

Notes :

1.

In addition to the two above, other formulae can be derived from them (textbook formulae 2.7 and 2.9). I do not believe in trying to memorize too many formulae, I think it is better to know the few fundamental ones, and derive other formulae and/or quantities as the need arises

2.

in formula 2.8 (the equivalent of my expression for x(t)), the book omits the term x₀. This is probably unwise, since it is not always possible to assume that x=0 when t=0, especially when facing problems dealing with more than one body. And in fact, in Example 11 (page 43) the book makes non-explicit use of a term x₀

Graphical interpretation : the expression giving velocity as a function of time is, as before, a straight line in the v-t plane. The (constant) slope of the line is the (constant) acceleration. The expression for x(t) is quadratic in t. In the simplest case x₀=v₀ = 0,the curve is a parabola symmetric around the origin. Remember how in the case of constant velocity, the velocity was given by the (constant) slope of the x vs. t line. This is still true even for the the non-straight case of constant acceleration : the velocity, which is continuously changing, is given by the local slope of the curve, i.e. the tangent to the curve at any given point.

In general, a parabola and a straight line will meet in two points (when they do meet). This means that problems involving the "meeting" of two bodies, one moving with constant velocity and one with constant acceleration will have two solutions,which might not always be real (see e.g. Example 11). Such problems are solved as before, i.e. finding the conditions for which the two bodies are at the same place at the same time.

Freely Falling Bodies

In order to study the motion of freely falling bodies, we only need to know one fact : a body falling freely under the effect of gravity experiences a constant acceleration "g" directed downward. g is known to be, on earth , about 9.80 m/s².

The fact that all bodies fall at the same rate, might be counter-intuitive at first thought (for many centuries, following Aristoteles ideas, "philosopers" would have sworn that heavier bodies fall faster). But on second thought, it is more logical to believe that all bodies fall at the same rate. Nice conceptual example ("gedanken experiment") by Galileo : take a heavy and a light ball, we are told that the light one will fall more slowly. But now attach them together with a very thin thread : the light ball, by its intrinsic lower rate of falling, should retard the fall of the heavy one...But,on the other side, the combination of the two balls is heavier than each single one, so they should fall even faster.... We are then facing an absurdity, and the way out of it is to assume that all bodies fall at the same rate. But, if you still have any doubt, you can just do some clever measurement. This is what Galileo did, and this is why he is considered the father of modern science: it is not just enough to have a nice theory, but you must verify whether it agrees with the data. If it doesn't, and your measurements were done right, then you must as well throw away your theory and go back to the drawing board....

But we are digressing, coming back to a free falling object, it is a fact that, by means of the formulae we have just learnt, we can completely describe its motion i.e. we can predict, at any instant its velocity and position (provided we know the initial conditions).

It is also useful to think of the graph of velocity vs. time of an object tossed up in the air...

Some results can be surprising at first, but with good reasoning everything makes sense. I find Example 18 particularly instructive :

a shotgun fires a pellet either straight up or straight down. In either case the magnitude of the initial velocity, as the pellet leaves the gun, is the same. In which case will the pellet hit the ground with higher velocity ?

To answer the question, we will find out the velocity of the upward going pellet when it comes back to its starting point.

The counter-intuitive result (the final velocity is the same) will be less surprising when we learn about Conservation of Energy, in a few weeks.