Chapter 2. Motion along a straight line

Chapter 2 Motion along a straight line

Introduction: Study of the motion of objects Physics studies: Properties of matter and energy: solid state physics, thermal physics/ thermodynamics, atomic physics, particle physics, quantum physics, new dark matter and dark energy. This chapter: Motion of objects The root of physics and modern science: Describing motion quantitatively is needed to study/understand the evolution/ dynamics! Interactions (forces): Four of them gravitational interaction, electromagnetic interaction, weak interaction, strong interaction. (Maybe more?) Evolution of systems the dynamics: classical dynamics, thermodynamics, quantum dynamics, quantum electrodynamics (QED), quantum chromodynamics (QCD),

There are moving objects all around us: rock, body, atoms, starts, and other trickier things like fields, photons, etc. The study of motion is called kinematics. Still or Moving? relative The Earth orbits around the Sun A roadway moves with Earth s rotation Everything moves

2.2 Motion The simplest case: motion that takes place in a straight line 1. How to describe motion, mathematically/quantitatively Important concepts: displacement, velocity, speed, acceleration, deceleration 2. How forces cause the change in motion: Force causes the objects to speed up, slow down, or maintain the same rate. 3. Assumptions for simplicity: The moving object here will be considered as a particle. If we deal with an extended object, we will assume that all particles on the body move in the same fashion (no spinning, no internal vibration, etc.)

2.3 How to describe motion: Position, displacement, and their dependence on time Position of an object: Coordinates in a scaled reference system Given a reference point, (the origin ). The positive direction: the direction the coordinates increases The negative direction: the direction the coordinates decreases -10 m 0 m 10 m Displacement: The change in the coordinates of the position of the body. If the x-coordinate of a body changes from x 1 (t 1 ) to x 2 (t 2 ), then the displacement, Δx = (x 2 -x 1 ). (Final position) (Initial position) It can be either positive or negative. -10 m 0 m 10 m

Displacement is a vector quantity. It has magnitude & direction information. A displacement is x = -10 m means that the object has moved towards decreasing x-axis by 10 m. + displacement -10 m 0 m 10 m - displacement -10 m 0 m 10 m

A common way to describe the motion of an object is to show a graph of its position as a function of time, the x-t graph

2.4 Average Velocity and Average Speed Average velocity, or v avg, is defined as the displacement over the time duration. v avg = Δx Δt = x 2 x 1 t 2 t 1 The average velocity has the same sign as the displacement: Because time always increases!

2.4 Average Velocity The magnitude of the slope of the x-t graph gives the average velocity Here, the average velocity between two points: is:

2.4 Average Speed displacement Average speed is the ratio of the total distance traveled to the total time duration. It is a scalar quantity, and does not carry any sense of direction. S avg = Total Dis. Δt Caution: Total distance may NOT be the same with the amplitude of the displacement! à S avg is not always = V avg - One-dimensional case - Two/Three-dimensional case

Total distance and the amplitude of the displacement: Several different scenarios

January 13

Displacement Δx = x 2 x 1 = x 2 (t 2 ) x 1 (t 1 ) Average velocity v avg = Δx Δt = x 2 x 1 t 2 t 1 Average speed S avg = Total Dist. Δt -10 m 0 m 10 m

Example, motion: Let s draw the motion

x 2.0 km Walking Gas station Driving 8.4 km O t 30 min We need to find out all the numbers on the t- and x-axis!

Example, motion: v avg = Δx Δt = x 2 x 1 t 2 t 1 0.12 h 0.50 h

Example, motion: Total Dist. S avg = Δt Units of time: minute à hour 0.12 h 0.50 h

What does the x-t graph look like now? Position x walking at pump driving time t

Position x walking at pump driving Walking??? 45 min time t

Position x walking at pump walking driving 45 min time t

2.5: Instantaneous Velocity and Speed

2.5: Instantaneous Velocity and Speed The instantaneous velocity of a particle at a particular instant is the velocity of the particle at that instant. This can be obtained by letting Δt approach 0 in: Δx v(t) = lim Δt 0 Δt = dx(t) dt 1. The instantaneous velocity is the slope of the tangent of the position-time graph at that particular instant of time. 2. Velocity is a vector quantity: amplitude + direction. 3. Instantaneous velocity could be very different from the average velocity

2.5: Instantaneous velocity: Direction and amplitude when not along a straight line D in time δt/3 D in time 2δt/3 Instantaneous velocity: v = lim δt 0 D δt What is the direction?

2.5: Instantaneous velocity: Direction and amplitude when not along a straight line V 1 at time t 1 V 2 at time t 2

2.5: Instantaneous Velocity and Speed Instantaneous speed s: the magnitude of instantaneous velocity v Δx s(t) = v(t) = lim Δt 0 Δt = dx(t) dt 1. +10 m/s and -10 m/s velocity have the same speed 10 m/s 2. Instantaneous speed could be very different from the average speed. Moving at a constant speed and moving at a constant velocity are different!

Derivatives??????????????????

Derivatives

Example, instantaneous velocity:

v(t) = lim Δt 0 Δx Δt = dx(t) dt

It looks like this! Why these are straight lines?? Acceleration & force

2.6: Average and instant accelerations Average acceleration is the change of velocity over the change of time, a avg = v(t 2) v(t 1 ) t 2 t 1 = Δv Δt The instantaneous acceleration is defined as: Δv(t) a(t) = lim Δt 0 Δt = dv(t) dt In terms of the position function, the acceleration can be defined as: a(t) = dv(t) dt = d dt dx(t) dt The SI units for acceleration are m/s 2. = d 2 x(t) dt 2

2.6: Average and instant accelerations Colonel J. P. Stapp in a rocket sled, which undergoes sudden change in velocities. If a particle has the same sign for velocity and nonzero acceleration, then that particle is speeding up. Conversely, if a particle has opposite signs for the velocity and non-zero acceleration, then the particle is slowing down. The magnitude of acceleration falling near the Earth s surface is 9.8 m/s 2, and is often referred to as g. Our bodies often react to accelerations but not to velocities: Riding a fast car feels very different from experiencing a sudden brake!

Example, acceleration: Origin Positive direction x (m)

Example, acceleration: Δx v(t) = lim Δt 0 Δt = dx(t) dt a(t) = dv(t) dt = d dt dx(t) dt = d 2 x(t) dt 2

Example, acceleration:

January 18

Last time

Example, acceleration: Motion is described by x(t), v(t), a(t). We need to discuss all three quantities, magnitude &direction!! x=4-27t+t 3 v=-27+3t 2 + a=6t

Example, acceleration:

Please try to describe the motion for t < 0 s yourself after class.

Section concept review by doing some exercises 2.4.2. Which one of the following statements concerning speed is true? a) Speed is always a positive number or zero. b) Speed can be a positive or negative number. c) Speed is always a negative number. d) The direction of the speed is directed from the starting point of motion to the ending point. e) The average speed is always the same as the instantaneous speed.

2.4.2. Which one of the following statements concerning speed is true? a) Speed is always a positive number or zero. b) Speed can be a positive or negative number. c) Speed is always a negative number. d) The direction of the speed is directed from the starting point of motion to the ending point. e) The average speed is always the same as the instantaneous speed.

2.4.4. A motorcycle travels due south covering a total distance of 80.0 kilometers in 60.0 minutes. Which one of the following statements concerning this situation is necessarily true? a) The velocity of the motorcycle is constant. b) The acceleration of the motorcycle must be non-zero. c) The motorcycle traveled 40.0 kilometers during the first 30.0 minutes. d) The speed of the motorcycle must be 80.0 kilometers per hour throughout the entire trip. e) The average velocity of the motorcycle is 80.0 kilometers per hour, due south.

2.5.1. Which one of the following position versus time graphs depicts an object moving with a negative constant velocity?

2.5.2. Which one of the following quantities can be determined from the slope of a position versus time graph for an object in motion? a) position b) velocity c) acceleration d) distance traveled e) displacement

2.5.3. Complete the following statement: For an object moving at constant velocity, the distance traveled a) increases for each second that the object moves. b) is the same regardless of the time that the object moves. c) is the same for each second that the object moves. d) cannot be determined, even if the elapsed time is known. e) decreases for each second that the object moves.

2.5.4. dog is walking along a street. As the dog moves, a graph is made of its position on the vertical axis with the elapsed time on the horizontal axis. The slope of the curve is determined at some point on the graph. The slope of this curve is a measurement of which of the following parameters? a) the dog s instantaneous velocity b) the dog s acceleration c) the dog s speed d) the dog s average velocity e) the elapsed time for the dog s walk

2.5.5. Starting from rest, a particle that is confined to move along a straight line is accelerated at a rate of 5.0 m/s 2. Which one of the following statements concerning the slope of the position versus time (x-t) graph for this particle is true? a) The slope has a constant value of 5.0 m/s. b) The slope has a constant value of 5.0 m/s 2. c) The slope is both constant and negative. d) The slope is not constant and increases with increasing time. e) The slope is not constant and decreases with increasing time.

2.6.7. A postal truck driver driving due east gently steps on her brake as she approaches an intersection to reduce the speed of the truck. What is the direction of the truck s acceleration, if any? a) There is no acceleration in this situation. b) due north c) due east d) due south e) due west

2.6.9. Which of the following situations is/are possible at a given time t? a) An object has an instantaneous velocity of 0 m/s and an acceleration of 0 m/s 2. b) An object has an instantaneous velocity of 0 m/s and an acceleration with a magnitude greater than 0 m/s 2. c) An object has an instantaneous velocity with a magnitude greater than 0 m/s and an acceleration of 0 m/s 2. d) Choices a, b, and c are all possible situations. e) Choices a, b, and c are not possible situations.

Motion: Displacement, speed/velocity, acceleration. Sometimes it is quite obvious: nearby, we can touch, straight line It is more complicated and often confusing at large distances!

Displacement, d(t) Velocity, v(t) 0 Acceleration a(t) 0 What caused all these? How? Force cause acceleration! The first thing: How to describe the motion with acceleration?

2.7: Constant acceleration along a straight line a(t) = dv(t) dt dv(t) = dv(t) = a(t)dt a(t)dt a(t)=a dv(t) = a dt v(t) = v(t 0 )+ a(t t 0 ) t 0=0, v(t) = v0 + at..(2-11) dx(t) dt = v(t) = v 0 + at dx(t) = (v 0 + at)dt dx(t) = (v 0 + at)dt x(t) x(t 0 ) = v 0 (t t 0 )+ 1 2 a(t 2 t 2 0 ) x(t) x 0 = v 0 t + 1 2 at 2..(2-15) Eliminating t from the Equations (2-11) and (2-15):..(2-16)

(2-11)..(2-15) Let s exercise on these: 1. Eliminating a from the Equations (2-11) and (2-15) to get: x - x 0 = (v 0 +v)t/2..(2-17) 2. Eliminating v 0 from the Equations (2-11) and (2-15) to get: x - x 0 = vt at 2 /2..(2-18)

Quick summary

Discussion: Math & Physics v 2 = v 0 2 + 2a(x x 0 ) v = ± v 0 2 + 2a(x x 0 )

Motion with a constant acceleration: Graphical illustration Differentiating position graphàvalues for velocity graph Integrating the velocity graph à values for position graph. Differentiating velocity graphàvalue for constant acceleration graph Integrating constant acceleration graph à values for velocity graph

Example, constant acceleration:

A B v 2 = v 2 0 + 2a(x x 0 ) A : 8 2 = v 2 0 + 2a(20 0) 64 = v 2 0 + 40a (1) B : 0 2 = v 2 0 + 2a(70 0) 0 = v 2 0 +140a (2) (1) (2) : 64 = 100a a = 0.64m / s 2 (2) : 0 = v 2 0 +140 ( 0.64) v 0 = 89.6 = 9.5m / s

Free-Falling Acceleration: am example of constant acceleration Free falling: With no external forces acting on them except for their weight. Use the constant acceleration model with a replaced by -g, where g = 9.8 m/s 2 for motion close to the Earth s surface. Experimental evidence: In vacuum, a feather and an apple will fall at the same rate Photo on the right.

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Example: constant acceleration 1-d motion: define an axis +: upward

What does it mean by reaching the maximum height? Knowing v 0, a, and v. Looking for t Which equation to use? v = v 0 + at at = v v 0 t = v v 0 a = 0 12m / s 9.8m / s 2 =1.2s

Knowing v 0, a, v, and t. Looking for (y-y 0 ) Which equation to use? (2-15): y y 0 = v 0 t + 1 2 at 2 y y 0 = (12m / s)(1.2s)+ 1 2 ( 9.8m / s2 )(1.2s) 2 y y 0 =14.4m 7.056m = 7.3m

(2.16): v 2 = v 0 2 + 2a(y y 0 ) 2a(y y 0 ) = v 2 v 0 2 y y 0 = v2 v 0 2a = 144m2 / s 2 19.6m / s 2 = 7.3m = 02 (12m / s) 2 2 ( 9.8)m / s 2

Knowing v 0, a=-g, and displacement y-y 0 =5.0m, looking for t (2.15): y y 0 = v 0 t + 1 2 at 2 Why two times?? 5.0m = (12m / s)(t)+ 1 2 ( 9.8m / s2 )(t) 2 5.0 =12t 4.9t 2 4.9t 2 12t + 5 = 0 t 1 = 0.53s,t 2 =1.9s

2-8: Non-constant acceleration: Graphical integration Starting from we obtain (v o = velocity at time t=0, and v 1 = velocity at time t = t 1 ). Note that Similarly, we obtain (x o = position at time t = 0, and x 1 = position at time t=t 1 ), and

Example, graphical solution:

(2.31)

B C

How does this cause serious consequence? v t = 7.2 km/h Only 5 ms! Force involved, to learn in later sections. v t = 7.2 km/h = 2 m/s, v h = 0 m/s 1 cm stretch between torso and head takes time t = (1cm)/(2m/s)= (0.01m)/(2m/s)=0.005 s = 5 ms

Discussion: How to avoid such incident?