Chapter 2 Describing Motion: Kinematics in One...

Chapter 2

Describing Motion: Kinematics

in One Dimension

• Introduction

• Reference Frames and Displacement

• Average Velocity

• Instantaneous Velocity

• Acceleration

(previous lecture)

(previous lecture)

(previous lecture)

(previous lecture)

(previous lecture)

• Acceleration

• Motion at Constant Acceleration

• Falling Objects

(previous lecture)

Recalling Last LectureRecalling Last Lecture

You need to define a reference frame in order to fully characterize the motion of an

object. In general we use Earth as reference frame for

Our measurements.

Displacement is how far an object is

from its initial position after an interval

of time ∆t:

(2.1)x x

Eq. 2.1 gives the magnitude of the vector

Average velocity is given by:

And average acceleration is:

Both velocity and acceleration are vectors (gives direction and magnitude) :

(2.1)x1 x2

(2.4)

(2.6)

Acceleration, velocity, displacement, and time.

Deceleration

The car from the previous example, now moving to the left and

decelerating.

Few Relevant NotesFew Relevant Notes

I. The magnitudes of the instantaneous velocity and instantaneous speed are the

same, though this is not necessarily true for average velocity and average speed.

II. If I give you an interval of time ∆t = t2 – t1 but do not specify the initial and final

time t1 and t2 , you can then assume t1 = 0 such that ∆t = t2 – t1 = t2 = t.

III. Similarly, if I give you a displacement ∆x = x2 – x1 but x1 is not specified, you can

assume x1 = 0 such that ∆x = x2 – x1 = x2 = x.

II and III are just a matter of redefining the origin of your coordinate system. For

Example:Example:

0 0x2x1X1 X2

Translate the origin from “0” to “x1” by applying a transformation:

X = x – x1 such that

X1 = 0 – x1 and X2 = x2 – x1 are the new set of coordinates.

Motion at Constant AccelerationMotion at Constant Acceleration

It is not unusual to have situations where the acceleration is constant. In this lecture

you will see one of such situations that you experience in your everyday life. But

before, let’s assume a motion in a straight line subject to a constant acceleration. In

this case the following is valid:

� The instantaneous acceleration and the average acceleration have the same

magnitude. We can then write:

(2.8)

We will use item II from the previous slide with following conventions:

t1 = 0 , t2 = t

and also define the position and velocity at t1 = t = 0 as

x1 = x0 , x = x2

v1 = v0 , v = v2

for our future developments.


With this in mind, let’s find some useful equations for situations with constant

acceleration that relate a, t, v, v0, x and x0 such that you can calculate any unknown

variable among them if you know the others.

Let’s first obtain an equation to calculate the velocity v of an object subjected to a

constant acceleration a:

Using the definitions introduced on the previous slide, we can rewrite eq. 2.6 as:

(2.9)

Eq. 2.10 can be rearranged:

Eq. 2.10 allows you to calculate the velocity v of an object at a given time t knowing

the its acceleration a and initial velocity v0.

(2.9)

(2.10)


Example: A car accelerates from rest at a constant rate of 5.0 m/s2. What is its

velocity after 5.0 s?

Solution:

Known variables:

v0 = 0.0 m/s

a = 5.0 m/s2

t = 5.0 st = 5.0 s

Using eq. 2.10:


We can also obtain an equation that gives the position x of an object knowing its

constant acceleration a, initial position x0 and initial velocity v0:

Using our definitions, eq. 2.4 can be written as:

(2.11)

On the other hand, since the velocity of the object changes from v0 to v at a constant

rate a, we have:

(2.12)


But (Eq. 2.11) = (Eq. 2.12). Then:

Using equation 2.10 in the above expression:

or

Eq. 2.13 allows you to calculate the position of an object at a given time t knowing

its acceleration a , initial velocity v0 and initial position x0 .

(2.13)


There is another very useful equation that applies to motion at constant acceleration

in situations where no time information has been provided.

We will use equations 2.10 and 2.13 to eliminate the time variable:

From eq. 2.10 �

(2.10)

(2.13)

From eq. 2.10 �

Using this expression in 2.13:

(2.14)


Example (Problem 28 from the text book): Determine the stopping distances for a

car with an initial speed of 95 Km/h and human reaction time of 1.0 s, for an

acceleration (a) a = -4.0 m/s2 ; (b) a= -8.8 m/s2 ?

Solution:

We want to convert from h to s. It is also wise to convert from Km to m:

(remember to carry some extra figures

through your calculations)through your calculations)

Next, we have to define our

reference frame (coordinate system)

in order to calculate the position

where the driver starts breaking.

Driver realizes he has to brake (v = v0)

Driver starts braking (v = v0)

cars stops (v = 0)

0 xx0


Continuing…………

Note that no information on the time needed to stop the car is provided. This problem

is better approached with eq. 2.14. The position where the brakes are applied can be

obtained if we observe that the car is moving at constant velocity v0 before the driver

reacts. We can then use the equation of motion for constant velocity to find x0:

As we have already mentioned, the final velocity v is zero in both (a) and (b) cases:

v (final) = 0.0 m/s

(continue on the next slide)


We have then the following initial conditions:

(a) The stopping distance for a = -4.0 m/s2 is:

(b) Similarly, the stopping distance for a = -8.0 m/s2 is:


Let’s summarize the set equations we just found:

(2.12)

(2.10)

Note that these equations are only valid when the acceleration is constant.

(2.14)

(2.13)


Example (Problem 21 from the text book): A car accelerates from 13 m/s to

25 m/s in 6.0 s. (a) What was its acceleration? (b) How far did it travel in this time?

Assume constant acceleration.

Solution:

This problem is about choosing an appropriate equation from the set of equations we

just found. For this purpose, first you have to write the parameters given in the

problem:

v0 = 13 m/s

v = 25 m/s

t = 6.0 s

(continue on the next slides)


(a) Here you want a knowing v0 , v and t. Which of the equations below you find more

appropriate to solve this problem?

(2.12)

(2.13)

(2.10)

It is clear that 2.10 is the better choice for this item.

;

(2.14)


(b) Here you want the distance travelled (x – x0) knowing v0 , v , t and a (calculated

in item (a) ). Now, which of the equations below you find more appropriate to solve

this problem?

(2.12)

(2.13)

(2.10)

Either equation 2.12 or 2.13 is a good choice. Let’s use 2.12:

(2.14)

(2.13)

FallingFalling ObjectsObjects

We know from observation that an object will fall if we through it from a bridge. It will

in fact gain speed as it falls. In other words, it is being accelerated.

This acceleration is due to the action of the Earth gravitational force upon the object.

We will come back to gravitation in few lectures from now.

We use the letter g to denote the gravitational acceleration.

� g is always different from zero at any position but the very center of the Earth.

We can say to very good approximation that g is constant:

(2.15)


Some Notes:

(a) It is also clear that certain objects fall faster

than others even though we release them from

rest (v=0) at the same time and from the same

height.

The reason for this is the air resistance.

In fact, all objects experience the same

gravitational acceleration g.

(b) Note that we dealing here with constant

constant acceleration.

Therefore all equations obtained on the previous slides

apply to this case.

(c) Since the motion will be vertical, we will use the y axis instead of the x axis.


Observing items (b) and (c) from the previous slide, we can rewrite the equations of

motion as:

(2.17)

(2.16)

Note: In most cases we will ignore the air resistance, unless otherwise stated.

(2.19)

(2.18)


Example (Problem 35 from the textbook): Estimate (a) how long it took King Kong

to fall straight down from the top of the Empire State Building (380 m high), and (b)

his velocity just before “landing”?

Solution:

Let’s chose downwards as the positive y direction (again, this is just a convention).

Again, we have to identify the initial conditions.

g = 9.80 m/s2

y - y0 = 380 m

v0 = 0.0 m/s

(continue on the next slides)

380 m

+y


(a) Here you want t knowing v0 , g and (y – y0) . Which of the equations below you

find more appropriate to solve this problem?

(2.17)

(2.19)

(2.18)

(2.16)

Equation 2.17 is more appropriated here:

Here, the minus sign is not physical since it would imply King Kong traveling back

time �

(2.19)


(b) Here you want v knowing v0 , g , (y – y0) and t. Which of the equations below you

find more appropriate to solve this problem?

(2.17)

(2.19)

(2.18)

(2.16)

Equation 2.16 is more appropriated here:

We might solve one more problem at the next class before starting chapter 3.

(2.19)

AssignmentAssignment 11

Textbook (Giancoli, 6th edition), Chapter 2:

1) Problem 5, page 39



4) Problem 47, page 414) Problem 47, page 41

Please, note the Questions and Problems are

two different things in the textbook. The

assignment above includes only Problems.

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