chapter 2 describing motion: kinematics in one...

Chapter 2

Describing Motion: Kinematics

in One Dimension

Dr. Mauricio Barbi

Email: [email protected]

Tel: 585-4260

Office: LB-212

Website:

http://ilc2.phys.uregina.ca/~barbi/academic/phys109/2010/phys109.html

Office Hours: Monday and Wednesday, 13:00-14:15h

Midterm exam: October 26, 13:00h-14:15h, CL-110

Final: December 14, 14:00h-17:00h, TBA

Office Hours and Exams

mailto:[email protected]

• Introduction

• Reference Frames and Displacement

• Average Velocity

• Instantaneous Velocity

• Acceleration

• Motion at Constant Acceleration

• Falling Objects

Motion is part of our daily life:

Get up in the morning

Walk to the UofR

Go for a lunch, etc.

If you are at rest and decide to go somewhere, you have

to “beat” your inertia and move; changing direction to

avoid a crazy teen driving a car also implies motion; etc.

Mechanics is the field in physics dedicated to the study

of motion. It is divided into two parts:

Kinematics: Describes how objects move (where an

object is at certain given time)

Dynamics: Describes the forces responsible for the

motion of objects (how they are set, and kept, in motion)

There are two types of motion: (a) Translation and (b)

Rotation.

We will start with one-dimensional translational motion:

objects that move along a straight-line path.

Introduction

Suppose I say:

“A person walks toward the front of a train at 5 km/h. The train is moving at 80 km/h”.

First question we should ask ourselves: The train is at 80 Km/h with respect to what?

Let’s assume it is with respect to the ground

Now, what is the speed of this person with respect to me if:

1) I am at rest on the train

2) I am at rest on the ground at a railway station

This example shows the importance

of identifying a frame of reference

(or reference frame) with respect to it

you can define your motion.

In our example, if you consider the person moving in the train:

- His speed is 5 Km/h if the reference frame is defined as the train;

- His speed is 85 Km/h if the reference frame is defined as the ground.

Reference Frames and Displacements

85 Km/h

5 Km/h

Ground

In the previous example, if I ask you the question: “Is the train going from Calgary to

Vancouver at 80 Km/h with respect to the ground?”, would you be able to give me an

answer?

No, because you do not know the direction it is moving (Calgary to Vancouver

or vice-versa?)

Despite the fact that we have identified the ground as the reference frame with

respect to it the speed of the train is measured, we have not yet fully characterized

the motion of the train: We also need to give its direction.

The direction can be specified by defining a

coordinate system which represents the

reference frame used to study the motion

of an object.

In our example, we can choose a

point along the railroad line as the origin, or

zero “0”, of our coordinate system.

We then draw two perpendicular axes, x and y,

crossing at the origin of the system.

But, how do we use such a system to define direction?


Definitions:

• Objects located to the right of the origin on the

x axis have positive x (+x) coordinates.

• Objects located to the left of the origin on the

x axis have negative x (-x) coordinates.

• Objects located above the origin on the y axis

have positive y (+y) coordinates.

• Objects located below the origin on the y axis

have negative y (-y) coordinates.

A point P in this coordinate system is fully identified

by a pair of coordinates (x,y)

For one-dimensional motion, we usually choose the x axis as the line of motion of an

object. If the motion is vertical, we usually choose the y axis.


P

(x,y)

x

y

Some notes:

The definitions of the signs on the previous slide

are arbitrary. You can choose your own sign

convention as long as you use it consistently

while solving a problem.

For example:

• you could have chosen positive x positions to

be located on the left side of the origin;

• or have chosen positive y positions to be

located below the origin.

We will see some examples later on.


+ y

- x

- y

+ x

- y

- x

+ y

+ x

Back to our example, let’s now locate the initial and final positions of the train in the

coordinate system we have just defined. Note that these two positions correspond to

measurements taken at two different times that we will call t1 and t2. Let’s then call

them positions x1 for t1 and x2 for t2.

Note: Positions x1 and x2 are measured relative to a point on the train (say, front of

the train) in our coordinate system.


x1 x2

x1 = Calgary

x2 = Vancouver

Now we know that the train is going

from x1 to x2, so we have a direction.

Say x1 represents Calgary and x2

represents Vancouver. You are now

able to tell me the direction the

train is travelling:

Calgary to Vancouver.

In the previous slide, the train has been displaced from its initial position x1 to its final

position x2.

We define this displacement as:

Note: Here the symbol Δ denotes “change”.


x1 x2

Note that Δx is not only a number

(scalar) but also carries information

about direction.

We can compile both information by

defining a vector:

Where the arrow on the top of Δx

defines a vector.

The magnitude of (2.2) is given by

(2.1).

(2.1)

(2.2)


Examples:

1) Suppose an object moves from position x1,

measured at an initial time t1, to position x2 at a

final time t2 as shown is figure (a). The blue arrow

represents the displacement x2 – x1, or the vector

Suppose x1 = 10 m and x2 = 30 m. Then, using

(2.1):

Δx is positive denoting a vector in the positive

x direction (right direction).

2) If we invert the initial and final positions such

that x1 = 30 m and x2 = 10 m as shown in figure

(b), then, again using (2.1):

Now Δx is negative denoting a vector in the

negative x direction (left direction).

(a)

(b)


Note:

You shall not confuse displacement with distance travelled.

Displacement gives how far distant an object has been displaced from its starting

point, along with the direction of such displacement. It is a vector quantity.

In the figure below, the starting position is at x1 = 0 m and the final position is at

x2 = 40 m. The displacement is:

regardless whether the object had to pass through

x3 = 70 m before reaching x2 = 40 m.

Distance gives the total length an object

travels from its starting position, regardless

the direction of its motion. It is a scalar quantity.

The total distance is given by: x1 x2 x3

Average Velocity

We usually make no distinction between the terms speed and velocity. However, in

physics they refer to different quantities and their distinction is very important.

Definitions:

Average Speed: Refers to how far an object travel in a given interval of time and is

related to total distance.

Speed is a scalar (do not depend on the direction of motion).

Example: In our previous example, an object moves

from x1 = 0 m to x2 = 40 m passing through x3 = 70 m

before reaching x2. Assuming the time elapsed to go

from x1 to x2 is given by Δt = t2 – t1 = 5 s, the average

speed is:

(2.3)

x1 x2 x3

Average Velocity

Average Velocity: Gives not only how fast an object travels but also its direction of

motion. Contrary to average speed, average velocity is related to displacement

instead of total distance traveled. Its magnitude is given by:

or

Note: The bar over v in (2.4) denotes “average”.

Velocity is a vector. The sign of Δx in (2.4) defines the direction of motion. If Δx < 0,

the average velocity is negative and therefore the object is moving toward negative x

coordinates, and vice-versa.

(2.4)

Average Velocity

x1 x2 x3

(a)

Examples:

1) Still using the example pictured in the figure (a),

an object is displaced from x1 to x2. Using

equation (2.4), the average velocity is:

The calculated average velocity is positive, therefore the object is moving toward

positive x coordinates.

Note that, for the given example, the calculated magnitude of the average speed

(20 m/s) is bigger than that for the average velocity (8 m/s).

Average Velocity

Examples:

2) The position of a runner as a function of time is plotted as moving along the x axis

of a coordinate system. During a 3.00 s time interval, the runner’s position

changes from x1 = 50 m to x2 = 30.5 m, as shown in figure (b). What was the

runner’s average speed? And its average velocity?

Average velocity:

Note: The absolute magnitude of both the average speed and average velocity

are the same. This is always true when the motion is all in one direction.

(b)

Instantaneous Velocity

So far we have used the term “average” for velocity and speed. This is due to the fact

that the velocity can change along the path of motion of an object, though we can

always assign an average velocity (or speed) to it knowing the time elapsed for the

object to go from an initial to a final position.

For example:

Figure (a) depicts a car moving at constant

velocity. In this particular case, the average

velocity is similar to the velocity of the car

at any instant of time.

On the other, figure (b) represents a car moving

at different velocities at different instant of time. In

this case, the average velocity is not necessarily

equal to the car velocity at a particular instant of

time.

a) constant velocity.

(b) varying velocity.

Instantaneous Velocity

We can use eq. (2.4) to define the velocity at any instant of time, or as we call:

instantaneous velocity. For this purpose, we can make the time interval very short.

In fact, we can make it so small that it will be close to zero. We say that Δt is tending

to zero, or that it is an infinitesimally short interval of time.

Note that I have dropped the bar on the top of v.

So, the magnitude of the instantaneous velocity

is represented by v in (2.5).

From now on, every time I use the term velocity

it will refer to the instantaneous velocity v instead

of the average velocity, unless otherwise stated.

a) constant velocity.

(b) varying velocity.

(2.5)

Δx

Acceleration

Acceleration gives how fast the velocity of an object is changing.

The average acceleration is defined as the change in velocity divided by the time

taken to make this change:

As velocity, acceleration is also a vector. It can have either positive or negative sign:

If v2 > v1 , Δv > 0 implies that the acceleration vector points towards positive x axis

If v2 < v1 , Δv < 0 implies that the acceleration vector points toward negative x axis

Note that acceleration has the units (SI) of m/s2.

(2.6)

Acceleration

An object can go faster (accelerate) or slow down (decelerate) depending on the

relative direction between the velocity and acceleration vectors.

The two examples below illustrated these concepts:

1) An car moves to the right of a coordinate system as shown in figure (a) (toward

positive x axis). The car slows down from an initial velocity v1 = 15.0 m/s to a final

velocity of v2 = 5.0 m/s in a time interval of 5.0 s. What is the car’s average

acceleration?

Using eq. 2.6,

Note that the acceleration is negative and points

in the opposite direction of the velocity.

Note also that if we invert the initial and final velocities such now v1 = 5.0 m/s and

v2 = 15.0 m/s, the resulting average acceleration will have the same magnitude but

with a positive sign pointing in the same direction as the velocity vector. Therefore, in

this case we have an acceleration.

Acceleration

Example: In this figure, the car is accelerated from a velocity v1 = 0 Km/h to v2 = 75

Km/h at a rate of 15 Km/h per second.

Acceleration

Similarly to instantaneous velocity, we can define instantaneous acceleration as:

.

(2.7)

chapter 2 describing motion: kinematics in one...

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