Chapter 2 Class Notes
This week, we begin the subject of kinematics, a mathematical description of how objects move in space and time. Several new concepts will be introduced in this chapter including
1. Vectors & Scalars
2. The four kinematic parameters
3. The three equations of kinematics.
4. Differential Calculus
Let’s begin with Vectors.
1. Vectors
In kinematics, vectors provide a mathematical way of describing the direction and length of a journey. Lets consider the example of the 100 mile journey from Prescott to Pheonix. As you drive down the hill, you find it’s a rather winding road, and when you get to Phoenix the odometer on your car will read 100 miles more than when you started, which is the distance travelled.
That same journey could be represented by an arrow, with the tip of the arrow located at Phoenix, and the tail of the arrow located at Prescott. The arrowhead shows the direction of the journey and the length of the arrow represents the displacement.
Displacement is different from distance, in that displacement depends only on the beginning and end points of the journey. The odometer in your car measures distance not displacement. You may think this a subtle difference until you consider the return journey, from Phoenix back to Prescott.
If you drive back up the hill, your odometer will add on another 100 miles, but if you had a displacement-meter in your car it would read zero! because your displacement from the point of origin, in this case Prescott, would be zero!
Displacements are unfamiliar to us because we do not have displacement meters in our cars, plus, we are not used to thinking of our location with respect to the origin of some grand coordinate system.
You may ask “Why use vectors like displacement at all?” The answer is because vectors are the easiest way mathematically to describe motion in time and space.
Vector Arithmetic
Vectors can be added together to find the resultant vector. There are two ways to do this; the graphical method, and the unit vector method. Lets start with the graphical method.
The Graphical Method of Adding Vectors
The graphical method, as the name suggests, requires drawing the vectors to scale on a piece of graph paper. There are homework exercises that illustrate this method, I’ll do Exercise 1 right now.
Exercise 1
a) You are given two vectors; A is 3 units long and B is 4 units long and you have to arrange them so that the resultant, lets call it vector C, is 7 units long.
Solution
Draw the vectors on a piece of graph paper as arrows. If you attach the tail of B to the tip of A you will get a vector that is 7 units long.
b) You are given two vectors A is 3 units long and B is 4 units long and you have to arrange them so that the resultant, lets call it vector C, is 1 unit long.
Solution
Draw the vectors on a piece of graph paper as arrows. If you flip vector A around and attach the tail of A to the tip of B you will get a resultant vector that is 1 unit long. Flipping A around like that is equivalent to writing -A, or multiplying A by -1.
c) You are given two vectors A is 3 units long and B is 4 units long and you have to arrange them so that the resultant, lets call it vector C, is 5 units long.
Solution The numbers 3, 4 and 5 should ring a bell. Ever heard of a 3,4 5 triangle? Draw A horizontally and B perpendicular to A, the hypotenuse of the resultant triangle is 5 units long.
The Unit Vector Method
The graphical method works fine for 2 dimensional vectors that can be drawn on a piece of paper. However, we live in a three dimensional (3D) universe and it is very difficult if not impossible to apply the graphical method to 3D vectors. In this situation we use the unit vector method. There are 3 unit vectors, each 1 unit long that point along the three orthogonal directions that are more familiarly known as x, y and z, but in the unit vector notation are called i, j and k. They simply define 3 directions in space.
To illustrate the use of unit vectors, I am going to repeat Exercise 1 using unit vectors.
a) Let vector A = 3 j and vector B = 4 j If I add them together, I get
A + B = 3 j + 4 j = 7 j which is a vector 7 units long pointing in the direction j
b) Let vector A = 3 j and vector B = 4 j If I subtract A from B I get
B - A = 4 j - 3 j = 1 j which is a vector 1 unit long pointing in the j direction.
c) Let vector A = 3 j and vector B = 4 k (note the direction change) If I add them together I get A + B = 3 j + 4 k which is the hypotenuse of the 3, 4, 5 triangle.
2. Introduction to One Dimensional Kinematics
By one dimensional, what I mean is objects moving from side to side or up and down.
Kinematic Variables
There are 4 kinematic variables, and they are displacement, velocity, acceleration and time. These variables are very important as they are the quantities that are substituted into the 3 equations of kinematics. We’ve already discussed displacement, so lets move on to velocity.
Velocity
Velocity is usually denoted by the vector v
Velocity is a vector because velocity = displacement / time, and displacement is a vector. The SI units of velocity are m/s.
Acceleration
Acceleration is usually denoted by the vector a
Acceleration is a vector because acceleration = velocity / time
and velocity is a vector, as already mentioned. The SI units of acceleration are m/s2.
Time
Time is a treated as a scalar, and it has units of seconds, s.
The 3 Equations of Kinematics
The easiest way to introduce the 3 equations of kinematics is by way of velocity - time graphs.
Consider the following velocity - time graph that depicts and object moving at constant velocity, v, for a time, t.
Interestingly, the area under the graph, which is the area of the rectangle vt is equal to the displacement. You can figure this out just based on the units, m/s x s = m.
Accelerating object
Next, consider the velocity - time graph for an object accelerating from some initial velocity vo to velocity v1. The displacement during the accelerating phase is the area under the graph, between times to and t1, the sum of one rectangular area and one triangular area.
The acceleration is the slope of the line of changing velocity. In terms of the notation on the graph,
a = v1 - vo
t1 - to
Now, if the final velocity, v1, is greater than the initial velocity, vo, the acceleration will come out positive. If however, the object were slowing down, the final velocity would be less than the initial velocity, and so the acceleration would actually come out negative. This is how we distinguish between objects speeding up and slowing down, by the sign of the acceleration vector.
The velocity, v, at any time, t, during the accelerating phase is just the equation of the sloping line,
v = vo + at
which is the first equation of kinematics!
The second equation of kinematics has to do with the displacement, x, during the accelerating phase.
Summing up the area of the small rectangle, vo (t1 - to), and the triangle (1/2 base x height), (v1 - vo)( t1 - to)/2. Added together yields
x = vo (t1 - to) + (v1 - vo)( t1 - to)/2
Now, lets pretend that to = 0, ie. the stop - watch starts at zero. Then we get
x = vo (t1) + (v1 - vo) (t1)/2
Next, using our first equation of kinematics, we can eliminate the velocity v1 since v1 = vo + a t1, yielding
x = vo (t1) + (a) (t1)2
2
This equation can be used to calculate the displacement, x(t), at any time, t, during the accelerating phase
so, x(t) = vo t + a t2
2
which is the second equation of kinematics!
The third equation of kinematics is
v2 - vo2 = 2 a x
which can be derived by eliminating the variable, t, between the first two equations. I will leave this as an exercise for the student.
Part of the difficulty in solving kinematics problems is trying to decide which of the three equations to use. You can make life easier for yourself by reading the question. Does the question give you the time or ask you for the time? If not, you can eliminate the first two equations from consideration right away since they both have time in them!
Lets do an example, the famous MOOSE problem! A driver is driving along at a constant velocity of 30 m/s when he suddenly spots a moose in the road 70m ahead. After a 0.5s reaction time, he steps on the brake and the car decelerates at 8 m/s2. The question is “does he hit the moose?”
Solution
This is a two part problem; a brief constant velocity phase before the driver steps on the brake and a decelerating phase after the driver steps on the brake. We need to calculate the distance covered in each of the two phases.
phase 1
x = vt
x = 30 x 0.5 = 15m
phase 2
The question does not give the time nor ask for it, so right away you know to use the 3rd equation of kinematics, which is
v2 - vo2 = 2 a x ( this is the physics )
0 - 302 = 2 (-8 ) x ( this is the math, note the minus sign )
x = 56.25m
Total distance travelled during phase 1 and phase 2 = 15 + 56.25m = 71.25m which is larger than the 70m stopping distance available. So, unfortunately, the moose dies!
Now, despite being a sad story, there’s an important detail to be learned about acceleration from this example, which is that we must add a minus sign to the acceleration if the object is slowing down.
Non - Uniform Acceleration
If an object is accelerating non-uniformly then the slope of the velocity - time graph is not constant, but is instead a wavy line.
The graph depicts an object that is repeatedly accelerating and decelerating, rather like someone who is a having trouble mastering the clutch on a manual drive car! Since the acceleration is constantly changing with time, the only reasonable way to talk about the acceleration is to specify the acceleration at a specific time, otherwise known as the instantaneous acceleration at a specific time.
We determine the acceleration the same way as we did before, simply by measuring the slope of the line in the velocity-time graph. Now, more specifically, we measure the slope of the line that is the tangent to the velocity curve at the time, t1, you are interested in knowing theacceleration, a.
It can be rather tedious to draw the tangent, and measure the slope, particularly if you have many accelerations to calculate. Fortunately, there is an entirely analytical way of doing it using differential calculus.
Differential Calculus
The math guys will take weeks to explain to you what really takes only a few minutes to understand. Differential calculus was developed by Sir Isaac Newton specifically to measure the slopes of graphs, amongst other things.
If you have a function v = t2 for example and you want to find the slope, you just differentiate the function. There is a little rule for differentiating functions and it goes like this, you bring the index, in this case 2, down in front of the t, and you subtract one power off what’s left. So you get the slope, or in this case, the acceleration = 2t. Now you just substitute the time, t, for which you need to know the acceleration.
The same trick works for displacement - time graphs, and what you get when you differentiate a displacement function, dx/dt, is the instantaneous velocity. The second derivative, d2x/dt2 , of the displacement function is equivalent to the first derivative of the velocity function, dv/dt, which, as already mentioned above is just the instantaneous acceleration. Homework Exercise 39 has to do with this topic.
The Importance of Diagrams
I cannot over emphasize the importance of diagrams for organizing your solution to a problem. In fact, I will insist that you always draw a diagram when solving physics problems. As an incentive, you will be given credit for drawing diagrams, on homework and exams. I’m not looking for a work of art, just a simple drawing that will help you summarize the given information.
I’ll illustrate what I mean with another example, this is the famous MOON problem! The question is “with what initial velocity do you have to launch a ball vertically to have it travel to a maximum height of 25m on the moon?”
Now, there is amazingly little information given in this problem, just one number, the height. The marvellous thing about physics though, is that you can figure out a lot from a little!
Let’s start with the diagram.
I have drawn the ball at three different times in it’s trajectory; the instant of release, the maximum height and the instant when the ball falls back down to the same height it started from.I’ve added a 1-D coordinate system, which I’m calling y. All displacements are measured up or down with respect to the origin 0. Additionally, I’ve added a sign convention in the top right hand corner to remind me that “up is positive” and “down is negative”, this is relevant for the vectors which I’ll discuss next.
I’ve identified two vectors, the velocity vector, v, and the acceleration vector, a. Now I’ll discuss the motion of the ball.
When the ball is thrown upward, it is being accelerated upward by the hand of the person launching it, but at the instant the ball is released, at y = 0, there is no longer any upward acceleration. In fact, once the ball is released, it is actually accelerating downwards, pulled down, as it is, by gravity. Now it is true that the ball is moving upwards as indicated by the velocity vector, but the ball is actually slowing down, so much so, that it actually stops for an instant at the top of it’s trajectory!
When the ball is at the top of it’s trajectory, the length of the velocity vector has shrunk to zero, indicating that the ball has stopped. The velocity vector, which was pointing upwards, flips over and points downward as the ball starts falling back down.
When the ball reaches y = 0, the velocity vector has grown to the same length as that with which it was launched, but the velocity vector is pointing down, not up. Notice that all through the trajectory, the acceleration has remained constant throughout at a = -g, the minus sign because gravity points downwards. Gravity is the only cause of the acceleration in this problem, there is no other source of acceleration whatsoever.
Once we have the diagram, next comes the physics.
Remember, we are trying to find vo given y = 25m. The best thing to do next is summarize what you know about the 4 kinematic parameters in a table, like this,
y = 25m ( maximum height )
yo = 0m ( initial height )
vo = ? (subscript o denotes “initial velocity at y = 0”)
v = 0 (velocity at y = 25m)
a = - g/6 (because we’re on the moon)
t = ?
Next, choose an equation of kinematics that is going to help you solve this problem. Now, the question does not give you any information about the time, and you are not asked for it, so right away you can zoom in on
v2 - vo2 = 2ay
We know that when y = 25m, v = 0, and so we have everything we need to solve for vo. Re-arrange the equation to make vo the subject,
vo = √(v2 - 2ay )
Lastly, the math, which means, plug in the numbers!
vo = √ ( 02 + 2 (9.8) 25 / 6 ) (Note: 2 minuses make a plus)
vo = 9 m/s
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