Motion in One Dimension

Motion in One Dimension Chapter 2

Displacement and velocity∙ ∙A → B • Displacement is the change of position of an object. • Displacement in SI units is measured in meters. • Coordinate systems are used to describe motion. • x axis for horizontal displacement • y axis for vertical displacement • Displacement → ∆x = xf – xi • Displacement = change in position = final position – initial position

Distance and displacement Distance and displacement are two quantities which may seem to mean the same thing, yet have distinctly different definitions and meanings. • Distance is a scalar quantity which refers to "how much ground an object has covered" during its motion. • Displacement is a vector quantity which refers to "how far out of place an object is"; it is the object's change in position. • To test your understanding of this distinction, consider the following motion depicted in the diagram below. A person walks 4 meters East, 2 meters South, 4 meters West, and finally 2 meters North. • Even though this person has walked a total distance of 12 meters, his displacement is 0 meters. During the course of his motion, he has "covered 12 meters of ground" (distance = 12 m). Yet when he is finished walking, he is not "out of place" - i.e., there is no displacement for his motion (displacement = 0 m). Displacement, being a vector quantity, must give attention and regard to direction. The 4 meters east is canceled by the 4 meters west; and the 2 meters south is canceled by the 2 meters north. 4 m 2 m 2 m 4 m

Displacement can be positive or negative Horizontal displacement (∆x) → Positive displacement ← Negative displacement Vertical displacement (∆y) ↑ Positive displacement ↓ Negative displacement

Velocity • Velocity is a vector quantity that measures how fast something moves from one point to another. • A vector quantity is a physical quantity that has both magnitude and direction. • A scalar quantity is a physical quantity that has only magnitude and no direction. • What is the difference between speed and velocity?

Average velocity is displacement in a given time • To calculate the average velocity of an object, you must know the object’s displacement. • You must also know the time the object left its initial position and the time it arrived at its final position. • The average velocity is equal to the displacement divided by the time during which the displacement occurred. • The SI unit of velocity is meters per second (m/s)

Average velocity v avg = __Δx_ Δt • Average velocity = change in position • change in time

??????? • P/1 If an arc of 60º on Circle I has the same length as an arc of 45º on Circle II, the ratio of the area of circle I to that of Circle II is: A) 19:9 B) 9:16 C) 4:3 D) 3:4

S1 = S2 π/ 3 r1 = π/4 r2 r1 = ¾ r2 A1 = π (r1)² A2 = π (r2)²

Average velocityExample # 1: During a race on level ground, Kelly covers 950 m in 2 min. while running due east. Find Kelly’s average velocity • Given: Δx = 950 m to the east • Δt = 2 min. • Unknown: v avg = ? • Solution: Use the average velocity equation: v avg = Δx / Δt v avg = 950 m / 2 min. v avg = 950 m / 120 s v avg = 7.92 m/s to the east

Physics Set # 3 August ____, 2008Student Name: _____________________________________________ Class Period: ______

Summer School Physics Set # 3 June 4, 2008Student Name: _____________________________________________ Class Period: ______

P/6 A bus traveled south along a straight path for 3.2 h with an average velocity of 88 km/h, stopped for 20 min, then traveled south for 2.8 h with an average velocity of 75 km/ha) What is the average velocity of the bus?B) What is the total displacement? V1 = 88 km/h ∆t 1 = 3.2 h 20 min V2 = 75 km/h ∆t2 = 2.8 h

Acceleration • Acceleration measures changes in velocity • Acceleration describes the rate of change of velocity in a given time interval. • The SI units for acceleration is meters per square second (m/s²) Average acceleration = change of velocity/change in time aavg= ∆v / ∆t = (vf – vi)/ (tf – ti)

Average accelerationExample # 1The Rockwell Bus from the Brownsville Urban System comes to a normal stop from 12 m/s to 0 m/s in 5 s. Find the average acceleration of the Bus. • Given: vi = 12 m/s vf = 0 m/s ∆t = 5 s • Unknown: a avg = ? • Solution: use the equation for average acceleration a avg = ∆v /∆t ∆v = vf – vi = 0 m/s – 12 m/s = - 12 m/s a avg = - 12 m/s / 5 s a avg = - 2.4 m/s²

Acceleration Velocity and acceleration • Acceleration has direction and magnitude • Acceleration is a vector quantity.

Summer School Physics Set # 4 June 4, 2008Student Name: ___________________________________ Class Period: ______

Displacement depends on acceleration, initial velocity and time v final v avg = ∆ x / ∆t For an object moving with constant acceleration, the average velocity is equal to the average of the initial velocity and the final velocity v avg = (vi+ vf) / 2 Displacement with constant acceleration: ∆ x = ½ (vi + vf) ∆t Velocity (m/s) v avg Time (s)

Final velocity depends on initial velocity, acceleration and time a avg. = (vf– vi) / ∆t (a avg)∙ ∆t = (vf – vi) Velocity with constant acceleration: vf= vi + (a avg)∙ ∆t Final velocity = initial velocity + (acceleration X time interval)

Final velocity depends on initial velocity, acceleration and time Displacement with constant acceleration: ∆x = ½ (vi + vf) ∆t Velocity with constant acceleration: vf= vi + (a avg)∙ ∆t Combining the two equations: ∆x = ½ [vi + (vi + (a avg)∙ ∆t) ] ∆t Displacement with constant acceleration ∆x = vi ∆t + ½ (a avg) (∆t )²

Extra Points P/1 Albert is a keen dog admirer and over the years has had a number of dogs. He has had an Alsatian, a Dalmatian, a Poodle and a Great Dane, but not necessarily in that order. Albert had Jamie first. The Dalmatian was an adored pet before the Great Dane. Sammy, the Alsatian, was the second dog Albert loved. Whitney was housed before the Poodle and Jimmy was not a Great Dane. Can you tell each of the dogs' name and the order in which Albert had them? Answer: Albert had Jamie the Dalmatian first, then Sammy the Alsatian, Whitney the Great Dane and finally, Jimmy the Poodle. P/2 A car traveled from London at a speed of 40mph. Its fuel consumption was 30 mpg. It had a 5 gallon tank which was full when it started, but at that very moment began to leak fuel. After 60 miles the car stopped with a completely empty tank. How many gallons per hour was it losing? Answer: 2 gallons per hour. The car traveled 60 miles before stopping, this means it used 2 gallons. The car took 1.5 hours to travel 60 miles. Therefore the car lost 3 gallons in 1.5 hours = 2 gallons per hour.

Set # 5 Answers P/1 A car with an initial speed of 23.7 km/h accelerates at a uniform rate of 0.92 m/s² for 3.6 s. Find the final speed and the displacement of the car during this time. • Given: vi = 23.7 km/h → vi = 6.58 m/s • a = 0.92 m/s² • ∆t = 3.6 s • Unknown: vf = ? , ∆x = ? • Solution:vf = a∙∆t + vi → vf = (0.92 m/s²) (3.6 s) + 6.58 m/s vf = 3.31 m/s + 6.58 m/s vf = 9.89 m/s ∆x = ½ (vi+vf)∙∆t ∆x = ½ (6.58 m/s + 9.89 m/s) (3.6 s) ∆x = 29.65 m

Set # 5 AnswersP/2A car with an initial speed of 4.30 m/s accelerates at the rate of 3 m/s².Find the final speed and the displacement after 5 s. • Given: vi = 4.30 m/s, a = 3 m/s² ∆t = 5 s • Unknown: vf = ?, ∆x = ?

Set # 5 AnswersP/3 A car can finish a 400 m race in 6 s. What is the minimum constant acceleration necessary? • Given: ∆x = 400 m • ∆t = 6 s • Unknown: a = ?

Set # 5 AnswersP/4 A car starts from rest and travels 5 s with a uniform acceleration of – 1.5 m/s².What is the final velocity of the car? How far does the car travel in this time interval? • Given: vi = 0 m/s, ∆t = 5s, a = - 1.5 m/s² • Unknown: vf = ?, ∆x = ?

Set # 5 AnswersP/5 A driver of a car traveling at – 15 m/s applies the brakes, causing a uniform acceleration of 2 m/s²If the brakes are applied for 2.5 s. What is the velocity of the car at the end of the braking period?. How far has the car moved during the braking period? • Given: vi = - 15 m/s, a = 2 m/s, ∆t = 2.5 s • Unknowns: vf = ? and ∆x = ?

Falling objects • Freely falling bodies undergo constant acceleration • In the absence of air resistance, all objects dropped near the surface of a planet fall with the same constant acceleration. Such motion is referred as free fall. • The free fall acceleration is denoted with the symbol g • At the surface of Earth the magnitude of g is approximately 9.81 m/s². • This acceleration is directly downward, toward the center of the Earth. • The downward direction is negative. • The acceleration of objects in free fall is g = - 9.81 m/s²

Introduction to Free Fall • A free-falling object is an object which is falling under the sole influence of gravity. Thus, any object which is moving and being acted upon only by the force of gravity is said to be "in a state of free fall." • This definition of free fall leads to two important characteristics about a free-falling object: 1.Free-falling objects do not encounter air resistance. 2. All free-falling objects on Earth accelerate downwards at a rate of approximately -9.81 m/s²

Free fall acceleration • This numerical value for the acceleration of a free-falling object is such an important value that it has been given a special name. • It is known as the acceleration of gravity – the acceleration for any object moving under the sole influence of gravity. As a matter of fact, this quantity known as the acceleration of gravity is such an important quantity that physicists have a special symbol to denote it – the symbol g. The value for g = - 9.81 m/s² • There are slight variations in this numerical value (to the second decimal place) which are dependent primarily upon altitude. • Acceleration is the rate at which an object changes its velocity. Between any two points in an object's path, acceleration is the ratio of velocity change to the time taken to make that change. • To accelerate at 9.81 m/s² means to change your velocity by 9.81 m/s each second.

If gravity had a value of -10 m/s² - - - - - - - -

Gravity is = – 9.81 m/s² • Free-falling objects are in a state of acceleration • Specifically, they are accelerating at a rate of -9.81 m/s². • This is to say that the velocity of a free-falling object is changing by 9.81 m/s every second. • If dropped from a position of rest, the object will be traveling: From: a = Δv / Δt solving for Δv Δv = a·Δt when t = 0 s → velocity = 0 m/sacceleration = - 9.81 m/s² t = 1 s v =(a·Δt ) =(-9.81 m/s²) (1 s) = - 9.81 m/sa = - 9.81 m/s² t = 2 sv = (a·Δt ) = (-9.81 m/s²) (2 s) = - 19.62 m/s a = - 9.81 m/s² t = 3 s v = (a·Δt ) = (-9.81 m/s²) (3 s) = - 29.43 m/sa = - 9.81 m/s² t = 4 s v =(a·Δt ) = (-9.81 m/s²) (4 s) = - 39.24 m/sa = - 9.81 m/s²

Free falling objectsStudent Name: _________________________ Sep/13/07

What goes up, must come down • When we throw an object up in the air, it will continue to move upward for a short time, stop momentarily at the peak, and then change direction and begin to fall. • Because the object changes direction, it may seem like the velocity and the acceleration are both changing. • Objects thrown into the air have a downward acceleration as soon as they are released. • When an object is thrown up in the air, there is an instant when the velocity is equal to 0 m/s. • Although the velocity is zero at the instant that the object reaches the peak, the acceleration is equal to – 9.81 m/s². • This acceleration is equal to – 9.81 m/s² at every instant regardless of the magnitude or direction of the velocity, • When an object is thrown up in the air it has a positive velocity and a negative acceleration, that means that the object is slowing down. • At the top of its path, the object’s velocity has decreased until it is zero. Although it is impossible to see this because it happens so quickly, the object is actually at rest for an instant, and even tough the velocity is zero at this point, the acceleration is still – 9.81 m/s². • The object then begins moving down, and when is moving down it has a negative velocity, and since the acceleration is still negative, the object will be speeding up.

Example # 1 A ball is thrown up in the air with an initial velocity of 44 m/s. Find the velocity and displacement of the ball in the first 10 s of this motion. Vi = 44 m/s

TimeVelocityAccelerationDisplacement 0 s0.00 m/s- 9.81 m/s²0.000 m 1 s34.19 m/s- 9.81 m/s²39.095 m 2 s 24.38 m/s- 9.81 m/s²68.380 m 3 s 14.57 m/s- 9.81 m/s² 87.855 m 4 s4.76 m/s - 9.81 m/s²97.520 m 5 s - 5.05 m/s - 9.81 m/s²97.375 m 6 s-14.86 m/s - 9.81 m/s²87.420 m 7 s-24.67 m/s- 9.81 m/s² 67.655 m 8 s-34.48 m/s- 9.81 m/s² 38.080 m 9 s-44.29 m/s- 9.81 m/s² - 1.305 m 10s-54.10 m/s- 9.81 m/s² - 50.500 m

What goes up, must come down V = _____ Vf = Vi =

Free FallExample # 1: Sofia hits a volleyball so that it moves with an initial velocity of 8 m/s straight upward. If the volleyball starts from 2 m above the floor, how long will it be in the air before it strikes the floor? Y axis Vi = 8 m/s X axis 2 m

Example # 1: Sofia hits a volleyball so that it moves with an initial velocity of 8 m/s straight upward. If the volleyball starts from 2 m above the floor, how long will it be in the air before it strikes the floor? • Given: vi = 8 m/s a = - 9.81 m/s² ∆y = - 2 m • Unknowns: vf = ? ∆t = ? • Solution: vf² = vi² + 2a∙∆y vf² = (8 m/s)² + 2 (- 9.81 m/s²) (- 2 m) vf² = 64 m²/s² + 39.24 m²/s² vf² = 103.24 m²/s²→ vf = √103.24 m²/s² vf = 10.16 m/s

Motion in One Dimension