PhysicsOur Physical World

Physics: Our Physical World

• Physics is relevant to us all. It allows us to analyze and appreciate our ever changing environment. • Problem solving in physics should be completed in a

systematic, organized and logical manner. It is necessary to have this structured approach when solving any type of problem in any of the sciences.

Evaluation

• Unit One: Forces and Motion (13.2%)• Unit Two: Conservation Laws (13.2%)• Unit Three: Modern Physics (13.2%)• Unit Four: Fields (13.2%)• Student Directed Study (7.2%)• Departmental (40%) – January!

Unit One: Forces and Motion (13%)

(15-20 classes)• PH30-FM1 Analyze motion in one- and two-dimensions,

including uniform motion, uniformly accelerated motion, circular motion, and projectile motion. • PH30-FM2 Analyze the effects of forces on objects

undergoing uniform motion, uniformly accelerated motion, circular motion, and projectile motion.

Forces and Motion 1 - Kinematics

• The branch of Physics that involves the description of MOTION without examining the forces which produce the motion.

• WHAT IS MOTION?• The movement of an object from one place to another • Measured over a time interval such as s, min, h• Relative to the position of the observer or the reference point.

• There are a variety of quantities associated with the motion of objects - displacement (and distance), velocity (and speed),acceleration, and time.

Scalars and Vectors

• Physics is a mathematical science.• The motion of objects can be described by words.• Words and phrases such as going fast, stopped, slowing

down, speeding up, and turning provide a sufficient vocabulary for describing the motion of objects.• We will be expanding upon this vocabulary list with words

such as distance, displacement, speed, velocity, and acceleration. 

Scalars and Vectors

• The mathematical quantities that are used to describe the motion of objects can be divided into two• Scalars are quantities that are fully described by a magnitude (or

numerical value) alone.• Vectors are quantities that are fully described by both a magnitude

and a direction.

Scalars and Vectors

• To test your understanding of this distinction, consider the following quantities. Categorize each quantity as being either a vector or a scalar.

Distance and Displacement

• Distance and displacement are two quantities that may seem to mean the same thing yet have distinctly different definitions and meanings.• Distance is a scalar quantity that refers to "how much ground an

object has covered" during its motion.• Displacement is a vector quantity that refers to "how far out of

place an object is"; it is the object's overall change in position.

Distance and Displacement

• To test your understanding of this distinction, consider the motion depicted in the diagram below. A physics teacher walks 4 meters East, 2 meters South, 4 meters West, and finally 2 meters North.

Distance and Displacement

• Even though the physics teacher has walked a total distance of 12 meters, her displacement is 0 meters. During the course of her motion, she has "covered 12 meters of ground" (distance = 12 m). Yet when she is finished walking, she is not "out of place" - i.e., there is no displacement for her motion (displacement = 0 m). Displacement, being a vector quantity, must give attention to direction. The 4 meters east cancels the 4 meters west; and the 2 meters south cancels the 2 meters north. Vector quantities such as displacement are direction aware. Scalar quantities such as distance are ignorant of direction. In determining the overall distance traveled by the physics teachers, the various directions of motion can be ignored.

Distance and Displacement

• Use the diagram to determine the resulting displacement and the distance traveled by the skier during these three minutes. • the skier moves from A to B to C to D.

Distance and Displacement

• The skier covers a distance of (180m + 140m + 100m) = 420mAnd has a displacement of 140m, rightward

Distance and Displacement

• What is the coach's resulting displacement and distance of travel?• the coach moves from position A to B to C to D.

Distance and Displacement

• The coach covers a distance of • (35 yds + 20 yds + 40 yds) = 95 yds• And has a displacement of 55 yards, left

Distance and Displacement

• To understand the distinction between distance and displacement, you must know the definitions. You must also know that avector quantity such as displacement is direction-aware and a scalar quantity such as distance is ignorant of direction. When an object changes its direction of motion, displacement takes this direction change into account; heading the opposite direction effectively begins to cancel whatever displacement there once was.

Distance and Displacement

• What is the displacement of the cross-country team if they begin at the school, run 10 miles and finish back at the school?• The displacement of the runners is 0 miles. While they have

covered a distance of 10 miles, they are not "out of place" or displaced. They finish where they started. Round-trip motions always have a displacement of 0.

Distance and Displacement

•  What is the distance and the displacement of the race car drivers in the Indy 500?• The displacement of the cars is somewhere near 0 miles

since they virtually finish where they started. Yet the successful cars have covered a distance of 500 miles.

Speed and Velocity

• Speed is a scalar quantity that refers to "how fast an object is moving." • Speed can be thought of as the rate at which an object

covers distance. • A fast-moving object has a high speed and covers a

relatively large distance in a short amount of time. Contrast this to a slow-moving object that has a low speed; it covers a relatively small amount of distance in the same amount of time. • An object with no movement at all has a zero speed.

Speed and Velocity

• Velocity is a vector quantity that refers to "the rate at which an object changes its position." • Imagine a person moving rapidly - one step forward and

one step back - always returning to the original starting position. While this might result in a frenzy of activity, it would result in a zero velocity. Because the person always returns to the original position, the motion would never result in a change in position. • Since velocity is defined as the rate at which the position

changes, this motion results in zero velocity.

Speed and Velocity

• Velocity is a vector quantity. As such, velocity is direction aware. • It would not be enough to say that an object has a velocity

of 55 mi/hr. For instance, you must describe an object's velocity as being 55 mi/hr, east. • This is one of the ESSENTIAL differences between speed

and velocity.

Speed and Velocity

• The task of describing the direction of the velocity vector is easy. The direction of the velocity vector is simply the same as the direction that an object is moving. It would not matter whether the object is speeding up or slowing down. If an object is moving rightwards, then its velocity is described as being rightwards. If an object is moving downwards, then its velocity is described as being downwards. So an airplane moving towards the west with a speed of 300 mi/hr has a velocity of 300 mi/hr, west. 

Speed and Velocity

• Calculating Average Speed and Average Velocity• As an object moves, it often undergoes changes in speed. For example,

during an average trip to school, there are many changes in speed. Rather than the speed-o-meter maintaining a steady reading, the needle constantly moves up and down to reflect the stopping and starting and the accelerating and decelerating. One instant, the car may be moving at 50 mi/hr and another instant, it might be stopped (i.e., 0 mi/hr). Yet during the trip to school the person might average 32 mi/hr.

• The average speed during an entire motion can be thought of as the average of all speedometer readings. If the speedometer readings could be collected at 1-second intervals (or 0.1-second intervals or ... ) and then averaged together, the average speed could be determined. Now that would be a lot of work. And fortunately, there is a shortcut.

Speed and Velocity

• Average vs. Instantaneous Speed

Speed and Velocity

• The average speed during the course of a motion is often computed using the following formula:

• In contrast, the average velocity is often computed using this formula

Speed and Velocity

• While on vacation, Lisa Carr traveled a total distance of 440 miles. Her trip took 8 hours. What was her average speed?• To compute her average speed, we simply divide the

distance of travel by the time of travel.

Speed and Velocity

• Lisa Carr averaged a speed of 55 miles per hour. She may not have been traveling at a constant speed of 55 mi/hr. She undoubtedly, was stopped at some instant in time (perhaps for a bathroom break or for lunch) and she probably was going 65 mi/hr at other instants in time. Yet, she averaged a speed of 55 miles per hour. 

Speed and Velocity

• Average Speed versus Instantaneous Speed• Instantaneous Speed - the speed at any given instant in time.• Average Speed - the average of all instantaneous speeds; found

simply by a distance/time ratio.

Speed and Velocity

• Moving objects don't always travel with erratic and changing speeds. Occasionally, an object will move at a steady rate with a constant speed.

• For instance, a cross-country runner might be running with a constant speed of 6 m/s in a straight line for several minutes. If her speed is constant, then the distance traveled every second is the same. The runner would cover a distance of 6 meters every second. If we could measure her position (distance from an arbitrary starting point) each second, then we would note that the position would be changing by 6 meters each second.

• This would be in stark contrast to an object that is changing its speed. An object with a changing speed would be moving a different distance each second

Speed and Velocity

Speed and Velocity

• Now let's consider the motion of that physics teacher again. The physics teacher walks 4 meters East, 2 meters South, 4 meters West, and finally 2 meters North. The entire motion lasted for 24 seconds. Determine the average speed and the average velocity.

Speed and Velocity

• The physics teacher walked a distance of 12 meters in 24 seconds; thus, her average speed was 0.50 m/s. • However, since her displacement is 0 meters, her average

velocity is 0 m/s. • Remember that the displacement refers to the change in

position and the velocity is based upon this position change. In this case of the teacher's motion, there is a position change of 0 meters and thus an average velocity of 0 m/s.

Speed and Velocity

• Use the diagram to determine the average speed and the average velocity of the skier during these three minutes. • the skier moves from A to B to C to D.

Speed and Velocity

• The skier has an average speed of • (420m)/(3min) = 140 m/min• And an average velocity of • (140m, right)/(3min) = 46.7 m/min, right

Speed and Velocity

• What is the coach's average speed and average velocity?• coach moves from position A to B to C to D.

.

Speed and Velocity

• Coach has an average speed of • (95 yd) / (10min) = 9.5 yd/min• And an average velocity of • (55 yd, left) / (10 min) = 5.5 yd/min, left

Acceleration

• Acceleration is a vector quantity that is defined as the rate at which an object changes its velocity. An object is accelerating if it is changing its velocity.

• Sports announcers will occasionally say that a person is accelerating if he/she is moving fast. Yet acceleration has nothing to do with going fast. A person can be moving very fast and still not be accelerating. Acceleration has to do with changing how fast an object is moving.

• If an object is not changing its velocity, then the object is not accelerating

• Animation and questions

Acceleration

• The Meaning of Constant Acceleration• Sometimes an accelerating object will change its velocity

by the same amount each second.• This is referred to as a constant acceleration• An object with a constant acceleration should not be

confused with an object with a constant velocity. Don't be fooled!

Acceleration

Acceleration

• If we were to observe the motion of a free-falling object (free fall motion will be discussed in detail later), we would observe that the object averages a velocity of approximately 5 m/s in the first second, approximately 15 m/s in the second second, approximately 25 m/s in the third second, approximately 35 m/s in the fourth second, etc.

TimeInterval

Velocity ChangeDuring Interval

Ave. VelocityDuring Interval

Distance Traveled

During Interval

Total Distance Traveled from0 s to End of

Interval0 – 1.0 s 0 to ~10 m/s ~5 m/s ~5 m ~5 m

1.0 – 2.0 s ~10 to 20 m/s ~15 m/s ~15 m ~20 m2.0 – 3.0 s ~20 to 30 m/s ~25 m/s ~25 m ~45 m3.0 – 4.0 s ~30 to 40 m/s ~35 m/s ~35 m ~80 m

Acceleration

• Calculating the Average Acceleration• The average acceleration (a) of any object over a given

interval of time (t) can be calculated using the equation

Acceleration

• If an object has an acceleration of 10 m/s/s.

• Otherwise known as 10 m/s2 

Acceleration

• Acceleration is a vector quantity• So…The direction of the acceleration vector depends on two things:

• whether the object is speeding up or slowing down• whether the object is moving in the + or – direction

• The general principle for determining the acceleration is:• If an object is slowing down, then its acceleration is in the opposite

direction of its motion.

Acceleration

• In Example A, the object is moving in the positive direction (i.e., has a positivevelocity) and is speeding up. • When an object is speeding up, the acceleration is in the

same direction as the velocity. Thus, this object has a positive acceleration.

Acceleration

• In Example B, the object is moving in the negative direction (i.e., has a negative velocity) and is slowing down. • When an object is slowing down, the acceleration is in the

opposite direction as the velocity. Thus, this object also has a positive acceleration.

Acceleration

• What if the acceleration of the object is in the negative direction?

.

Acceleration

• In Example C, the object is moving in the positive direction (i.e., has a positive velocity) and is slowing down. • When an object is slowing down, the acceleration is in the

opposite direction as the velocity. Thus, this object has a negative acceleration.

Acceleration

• In Example D, the object is moving in the negative direction (i.e., has a negativevelocity) and is speeding up. • When an object is speeding up, the acceleration is in the

same direction as the velocity. Thus, this object also has a negative acceleration.

Acceleration

• Animation

Acceleration

• Check Your Understanding; Remember FORMULA!

Acceleration

• Practice A – 2m/s2 (positive acceleration)• Practice B – -2m/s2 (negative acceleration)

Computer Activity…

• Name that Motion - Interactive link in Physics Classroom• 11 animations and make an effort to match each animation

to a verbal description

Uniform Motion

• simplest form of MOTION• Following a STRAIGHT path at a CONSTANT speed• Motion with a CONSTANT velocity

Uniform Motion - Examples

• EXAMPLES:• Cruising Airplane• Car moving on a STRAIGHT road at a CONSTANT speedNon Example:• Driving on a hilly or curvy road• School bus trip

Uniform Motion - Formula and Variables

• v = d/∆t• v – velocity (easiest to measure in m/s)• d – displacement• ∆t – change in time

Uniform Motion – Constant speed

• Constant Speed - Constant speed occurs when the object travels the same distance in equal periods of time. It is an example of uniform motion.• THEREFORE: Constant speed = Uniform motion

Uniform Motion - Equation Example

EXAMPLE The car travels at a rate of 2.0 m every 0.1 s. Since the car travels the same distance in equal periods of time the speed is said to be constant. If the car goes 12m in 0.6s what is the constant speed?FORMULAv = d/∆t

Uniform Motion - Answer

• 12.0m/0.6s = 20m/s

Uniformly Accelerated Motion (UAM)

• Uniform or CONSTANT acceleration is a type of motion in which the velocity of an object changes by an equal amount in every equal time period.• What is acceleration?• Acceleration, in physics, is the rate of change of velocity of

an object. An object's acceleration is the net result of any and all forces acting on the object, as described by Newton's Second Law. The SI unit for acceleration is the metre per second squared (m/s2).• An object is accelerating if it is changing its velocity.

UAM - Examples

• EXAMPLES• A ball running down an incline• A person falling from a plane• A bicycle on which you have applied the brakes• A ball dropped from the top of a ladder

• (Technically, because of friction and a non-constant gravitational field, etc., they are not quite Uniformly Accelerated Motion, however, at this point we will treat them as if they are, because it is close enough, for now.)

UAM - Formulas

• These are the equations that describe an object in Uniformly Accelerated Motion:• vf = vi + aΔt• Δd = vi Δt + ½ (aΔt2)• vf 2 = vi 2 + 2aΔd• Δd = 1/2 (vi + vf ) Δt

• (OTHERWISE KNOWN AS KINEMATIC EQUATIONS)

UAM - Variables

• vi = velocity initial (easiest to measure in m/s)• vf = velocity final (easiest to measure in m/s)• a = acceleration • Δd = displacement • Δt = change in time

UAM – How it works…

• There are FIVE variables in the UAM equations. • There are FOUR UAM equations.• If you know THREE of the variables, you can determine the

other TWO variables.• This leaves you with ONE happy physics student. (note:

not one answer. There can be more than one answer.)

UAM- Equation Example

• Ms. Yedersberger is riding her bike at 22.9 km/h when she applies the brakes causing the bike to slow down with a constant acceleration.

• CONSTANT ACCELERATION – CAN use UAM FORMULAS!

• 3 UAM variables – not necessarily UAM formulas…

UAM- Equation Example

• Ms. Yedersberger is riding her bike at 22.9 km/h when she applies the brakes causing the bike to slow down with a constant acceleration. • After 1.01 seconds she has traveled 4.00m.A) What was her acceleration?B) What was her final speed?

UAM- Equation Example

• FIRST – IDENTIFY VARIABLES YOU HAVE!!!

UAM- Equation Example

• Ms. Yedersberger is riding her bike at 22.9 km/h when she applies the brakes causing the bike to slow down with a constant acceleration. • After 1.01 seconds she has traveled 4.00m.• vi = velocity initial 22.9 km/h• vf = velocity final ? – answer in B)• a = acceleration ? – answer in A)• Δd = displacement 4.00 m• Δt = change in time 1.01 s

UAM- Which Formula for A) acceleration?

HAVE:• vi = 22.9 km/h• Δd = 4.00 m• Δt = 1.01 sWANT:• a = ?DON’T HAVE:• vf = ?

FORMULAS• vf = vi + aΔt• Δd = vi Δt + ½ (aΔt2)• vf 2 = vi 2 + 2aΔd• Δd = 1/2 (vi + vf ) Δt

UAM - A) ACCELERATION

What Formula?• Δd = vi Δt + ½ (aΔt2)Rearrange so a = ?• Δd = viΔt + ½(aΔt2)• Δd − viΔt = 0.5(aΔt2)• (Δd − viΔt)/(0.5Δt2) = a• OR• a = (Δd − viΔt)/(0.5Δt2)

UAM - A) ACCELERATION

a = (Δd − viΔt)/(0.5Δt2) Remember:• vi = 22.9 km/h• Δd = 4.00 m• Δt = 1.01 sAny problems??? *** Convert to meters and seconds22.9 km/h * 1h/3600s * 1000m/1kmvi = 6.3611111111 m/s

UAM – A) ACCELERATION

PLUG IT IN.. • a = (Δd − viΔt)/(0.5Δt2) Variables• vi = 6.3611111111 m/s• Δd = 4.00 m• Δt = 1.01 s

a = (Δd − viΔt)/(0.5Δt2) a = [4.00m – (6.361m/s)(1.01s)] / [0.5(1.01)2]a = [4.00m – 6.42m ] / [0.5(1.02s2)]a = [-2.42472222222222222m] / [.51005s2]a = -4.753891230707 m/s2

a = -4.75 m/s2

UAM- Which Formula for B) final speed?

HAVE:• vi = 6.3611111111 m/s• Δd = 4.00 m• Δt = 1.01 s• a = = -4.753891230707 m/s2

(use unrounded answer)WANT:• vf = ?

FORMULAS• vf = vi + aΔt• Δd = vi Δt + ½ (aΔt2)• vf 2 = vi 2 + 2aΔd• Δd = 1/2 (vi + vf ) Δt

• CAN USE 3 FORMULAS – good for a check..

The reason there are 3 equations we could use is because after we have solved part (a) we now know four of the UAM variables and not just 3.

UAM- Which Formula for B) final speed?

HAVE:• vi = 6.3611111111 m/s• Δd = 4.00 m• Δt = 1.01 s• a = = -4.753891230707 m/s2

(use unrounded answer)WANT:• vf = ?

FORMULAS – Easier first!• vf = vi + aΔt• vf = 6.3611111111 m/s +

(-4.753891230707 m/s2 )(1.01 s)• vf = 6.3611111111 m/s +

(-4.80143014301407 m/s)• vf = 1.55968096809593 m/s

• vf = 1.56 m/s

Assignment

• Forces and Motion Assignment #1

Forces and Motion - Assignment #1

1) An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.

2) A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.