普通物理－直線運動

普通物理

Lecture 3

Motion in a Straight Line

Motion in a Straight Line

Contents

Kinematics 運動學

Quantities in Motion 運動三要素

Distance & Displacement 距離與位移

Distance & Displacement 距離與位移

Speed & Velocity 速率與速度

Average Velocity & Instantaneous Velocity

平均速度與瞬時速度

Accelerations ( average uniform &instantaneous)

平均、單一與瞬時加速度 Graphical Interpretation 直線運動圖解法 Kinematic Equations 運動方程式 2 Kinematic Equations 運動方程式 Free Fall 自由落體

Kinematics

 i i i i

The branch of physics involving the motion of an object and the relationship between that motion and other physics concepts

Kinematics is a part of dynamicsp y

In kinematics, you are interested in the , y description of motion 現象的描述

Not concerned with the cause of the

Not concerned with the cause of the motion

Quantities in Motion

Any motion involves three concepts

Displacement 位移

Velocity 速度

Velocity 速度

Acceleration 加速度

These concepts can be used to study objects in motion

in motion

Brief History of Motion

y

 i

Sumerian and Egypt

Mainly motion of heavenly bodiesy y

Greeks Greeks

Also to understand the motion of heavenly bodies

Systematic and detailed studiesy

“Modern” Ideas of Motion

 i

Copernicus

Developed the heliocentric systemp y

GalileoGalileo

Made astronomical observations with a telescope

telescope

Experimental evidence for description of ti

motion

Quantitative study of motion

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Position

motion motion

Displacement

p

 i i i i

Defined as the change in position

  x x_f  x_i

f stands for final and i stands for

initial

f i

initial

May be represented as y if vertical



Units are meters (m) in SI, centimeters (cm) in cgs or feet (ft) in US Customary

Vector and Scalar Quantities

 i i i i

Vector quantities need both magnitude (size) and direction to completely describe them

Generally denoted by boldfaced type and an arrow over the letter

+ or – sign is sufficient for this chapter

Scalar q antities are completel described Scalar quantities are completely described

by magnitude only

Displacement Isn’t Distance

p

 The displacement of an object is not the same as the distance it travels

 Example: Throw a ball straight up and then catch it at the same point you released itp y

The distance is twice the height

The distance is twice the height

Speed

p

 i i

The average speed of an object is defined as the total distance traveled divided by the

total time elapsed

 total distance Average speed total time  d v t

Speed is a scalar quantity

t

12

Speed

速率

p

 Average speed 速率 totally ignores any variations

in the object’s actual motion during the trip

Average Speed must be positive

Th t t l di t d th t t l ti ll

g p p

The total distance and the total time are all that is important

Velocity

y

 It takes time for an object to undergo a displacement

 The average velocity is rate at which the displacement g y p occurs





x

x



x











x

x

x

v

t

t

t

 generally use a time interval, so let t _i = 0



t

t

t

Velocity

y

 Direction will be the same as the direction of the displacement (time d ec o o e d sp ce e ( e interval is always positive)

 + or - is sufficient

 + or - is sufficient

 Units of velocity are m/s (SI), cm/s

(cgs) or ft/s (US Cust ) (cgs) or ft/s (US Cust.)

 Other units may be given in a problem, but generally will need to be converted to these

Speed vs. Velocity

p

y

Cars on both paths have the same averageCars on both paths have the same average velocity since they had the same

displacement in the same time interval displacement in the same time interval

The car on the blue path will have a greater a erage speed since the distance it tra eled

16

average speed since the distance it traveled is larger

Graphical Interpretation of Velocity

p

p

y

 i i i i

Velocity can be determined from a position-time graph

Average velocity equals the slope 斜率of the line joining the initial and final positionsj g p

An object moving with a constant velocity

will have a graph that is a straight line will have a graph that is a straight line

Graphical Interpretation of Velocity

On an x versus t plot we can determine v_avg from the slope of the

p

p

y

straight line that connects point ( t₁ , x₁) with point ( t₂ , x₂ ). In the

plot below t₁=1 s, and t₂ = 4 s. The corresponding positions are: x₁ = 4 m and x = 2 m = – 4 m and x₂ = 2 m 2 1 2 1 2 ( 4) 6 m 2 m/s 4 1 3 s avg x x v t t          18

Average Velocity, Constant

g

y

The straight line g indicates constant

velocityy

The slope of the line is the value of line is the value of the average

velocity velocity

Average Velocity, Non Constant

g

y

The motion is non

The motion is non-constant velocity 

The average

velocity is the slope of the blue line

joining two points

Example 1:

An automobile travels on a straight road for 40 km at 30 km/h. If then continues in the same direction for another 40 km at 60 km/h then continues in the same direction for another 40 km at 60 km/h. (a)What is the average velocity of the car during this 80 km trip?

(b) Graph x-t and indicate how the average velocity is found on the graph.

t₁  (40 km)  1 33. (30 km / h) h. t2 40 0 67  ( km)  . (60 km / h) h. km x₁  40  x₂  40km (a) (30 km / h) (60 km / h) h km x x x v_avg   total   1   2  40 40  40 / t t t_total avg 67 . 0 33 . 1 2 1   (b)

Instantaneous Velocity

瞬時速度

y

 i i i i

瞬時速度

The limit of the average velocity as the time interval becomes infinitesimally short, or as the time interval approaches zero

x







x

v

t

The instantaneous velocity indicates what is

t

Instantaneous Velocity on a Graph

y

p

 i i i

The slope of the line tangent to the position-vs.-time graph is defined to be the

instantaneous velocity at that time

The instantaneous speed is defined as the magnitude of the instantaneous velocity magnitude of the instantaneous velocity

Uniform Velocity

y

 i i i i

Uniform velocity is constant velocity

The instantaneous velocities are alwaysy the same

All the instantaneous velocities will also

All the instantaneous velocities will also equal to the average velocity

Acceleration

 i i i

Changing velocity (non-uniform) means an acceleration is present

Acceleration is the rate of change of the velocityy       f i v v v a t t t

Units are m/s² (SI), cm/s² (cgs), and ft/s² (US t t_f t_i

Average Acceleration

g

When the signg of the velocity and the y

acceleration are the same (either positive or negativeg ), then the), speed is increasingp g

When the sign of the velocity and the

acceleration are in the opposite directions acceleration are in the opposite directions, the speed is decreasing

Instantaneous and Uniform

Acceleration

瞬時加速度與等加速度

Acceleration

 i i i

瞬時加速度與等加速度

The limit of the average acceleration as the

time interval goes to zero

When the instantaneous accelerations are alwaysy the same, the acceleration will be, uniform

The instantaneous accelerations will all

The instantaneous accelerations will all be equal to the average acceleration

Graphical Interpretation of

Acceleration

Acceleration

 A i

 Average acceleration

is the slope of the line

ti th i iti l

connecting the initial and final velocities on

l it ti h

a velocity-time graph

 Instantaneous

l i i h

acceleration is the

slope of the tangent to

th f th

the curve of the

velocity-time graph

Relationship Between Acceleration

and Velocity

and Velocity

Uniform velocity (shown by red arrowsUniform velocity (shown by red arrows maintaining the same size)

Acceleration equals zero Acceleration equals zero

Relationship Between Velocity and

Acceleration

Acceleration

 Velocity and acceleration are iny the same direction

 Acceleration is uniform (blue arrows maintain the

same length) same length)

 Velocity is increasing (red arrows are getting

longer)

30

longer)

Relationship Between Velocity and

Acceleration

Acceleration

 Acceleration and velocity are in opposite directions

 Acceleration is uniform (blue arrows maintain the

same length)

 Velocity is decreasing (red arrows are getting

Motion Diagram

Which one is not possible? Which one is not possible?

Example 2:

(a)If the position of a particle is given by x=20t-5t3_{, where x}

is in meters and t is in seconds, when, if, ever, is the particle’s velocity zero?

(b) When is its acceleration zero? (b) When is its acceleration zero?

(c) For what time range (+ or -) is acceleration negative? (d) Positive?

( )

t  20 15/  1 2. s

(a) v=x’=20-15t2₌₀

(b) F 0 30 fi d (0) 0 (b) From 0 = – 30t, we find a(0) = 0

(that is, it vanishes at t = 0)

( ) It i l th t ( ) 30 i ti f 0 (c) It is clear that a(t) = – 30t is negative for t > 0 (d) The acceleration a(t) = – 30t is positive for t < 0.

Kinematic Equations

運動方程式

q

 U d i i i i h if l i

運動方程式

 Used in situations with uniform acceleration

t





v

v

at

_

_





₂

v

v

a x





2



v

v

a x

Graphical Interpretation of the

Equation

Notes on the equations

q





v

v

at

Use when you don’t Use when you don’t

know and aren’t asked

t fi d th di l t

to find the displacement

Notes on the equations

q

t

2

v

v

t

v

x





















2









 Gives displacement as a function of velocity

and time and time

 Use when you don’t know and aren’t asked

for the acceleration for the acceleration

Notes on the equations

q

at

1

t

v

x





at

2

t

v

x









Gives displacement as a function of

time,

velocity

and

acceleration

velocity

and

acceleration



Use when you don’t know and aren’t

asked to find the

final velocity

Notes on the equations

q







2 2

₂

v

v

a x

 Gives velocity as a function of acceleration and





2



v

v

a x

y

displacement

 Use when you don’t know and aren’t asked for y

Problem-Solving Hints

g



Read the problem Draw a diagramg

Choose a coordinate system, label initial and final points indicate a positive

and final points, indicate a positive

direction for velocities and accelerations

Label all q antities be s re all the nits are Label all quantities, be sure all the units are

consistent

Convert if necessary

Choose the appropriate kinematic equation

40

Problem-Solving Hints, cont

g



Solve for the unknowns

You may have to solve two equations for y q two unknowns

Check your results Check your results

Estimate and compare



Example 3:

An electron with an initial velocity v₀=1.50 × 105 m/s enter a region of length L=1 cm where it is electrically accelerated. It emerges with v=5.70 × 106 m/s. What is its acceleration, assumed constant?

assumed constant?

v



v



2

a x

(



x

)

2 2 5 2 5 2

with x₀ = 0 and x = 0.010 m. Thus,

2 2 5 2 5 2 15 2 0 (5.7 10 m/s) (1.5 10 m/s) _{1.62 10 m/s .} 2 2(0.010 m) v v a x         42

Galileo Galilei

 1564 - 1642

 Galileo formulated the laws that

govern the motion of objects in free fall

 Also looked at:

 Inclined planes 斜面運動  Inclined planes 斜面運動  Relative motion 相對運動  Th t 溫度計  Thermometers 溫度計  Pendulum 鐘擺運動

Free Fall

 i

All objects moving under the

influence of gravity only are said to be in free fall

Free fall does not depend on p the object’s original motion

All objects falling near the All objects falling near the

earth’s surface fall with a

constant acceleration constant acceleration

The acceleration is called the

l ti d t it d

44

acceleration due to gravity, and

indicated by g 重量不同的兩物體由相同

Acceleration due to Gravity

y

 i

Symbolized by g g = 9.80 m/s²g

When estimating, use g ≡ 10 m/s2

g is always directed downward g is always directed downward

toward the center of the earth

Ignoring air resistance and assuming g doesn’t vary with altitude over short y

vertical distances, free fall is constantly accelerated motion

Free Fall – an object dropped

j

pp

 i i i i

Initial velocity is zero Let up be positivep p

Use the kinematic equations

Generally use y instead of

Generally use y instead of x since vertical  2 v_o= 0 a = g Acceleration is g = -9.80 m/s2 a g 46

Free Fall – an object thrown downward

Free Fall -- object thrown upward

j

p

Thrown upward

p

 i i

The motion may be symmetrical

Then t_upup = t_downdown

Then v = -v_o

The motion may not be symmetrical The motion may not be symmetrical

Break the motion into various parts

Non-symmetrical

Free Fall

Non-symmetrical

Free Fall

Free Fall

Free Fall

Need to divide the motion into segmentsg

Possibilities include

Upward and downward

Upward and downward portions

Th i l i

The symmetrical portion back to the release point and then the

non-symmetrical portion

Example 4:

A bolt is dropped from a bridge under construction, falling 90 m to the valley below the bridge

valley below the bridge.

(a) In how much time does it pass through the last 20% of its fall? (b) What is its speed, when it begins that last 20% of its fall

(c) What is its speed when it reaches the valley beneath the bridge? (c) What is its speed, when it reaches the valley beneath the bridge?

m gt gt y y 90 2 1 2 1 _{2 2} 0      

Upwards is chosen as the positive y direction _{y y g g}

2 2 0 sec 29 . 4 8 . 9 90 2    t

The time for the full fall is

The first 80% of its free fall distance is given by –72 = –g t2_{/2, which requires time}

t = 3.83 s.

(a) Thus, the final 20% of its fall takes t – t = 0.45 s.

(b) We can find that speed using v = -gt =-9.8 p g g × 3.83= -37.5 Therefore, |v| = 37.5 m/s, approximately.

Assignment 2

D e Date: 10/14/2009

Assignment 2

Due Date: 10/14/2009

1.

Julie drives 100 mi to Grandmother’s house. On the way to Grandmother’s, Julie drives half the distance at 40 mph and half the distance at 60 mph On her return trip she drives half the distance at 40 mph and half the distance at 60 mph. On her return trip, she drives half the time at 40 mph and half the time at 60 mph.

a. What is Julie’s average speed on the way to Grandmother’s house? b What is her average speed on the return trip?

b. What is her average speed on the return trip? 2.

A jet plane is cruising at 300 m/s when suddenly the pilot turns the engines up to full A jet plane is cruising at 300 m/s when suddenly the pilot turns the engines up to full throttle. After traveling 4.0 km, the jet is moving with a speed of 400 m/s.

a. What is the jet’s acceleration, Assuming it to be a constant acceleration? b. Is your answer reasonable? Explain.y p

3.

A student standing on the ground throws a ball straight up. The ball leaves the student’s

52

hand with a speed of 15 m/s when the hand is 2.0 m above the ground. How long is the ball in the air before it hits the ground?

Assignment 2

4.

A ball rolls along the frictionless track shown in figure below. Each segment of the track is A ball rolls along the frictionless track shown in figure below. Each segment of the track is

straight, and the ball passes smoothly from one segment to the next without changing speed or leaving the track. Draw three vertically stacked graphs showing position,

velocity, and acceleration versus time. Each graph should have the same time axis, and y, g p , the proportions of the graph should be qualitatively correct. Assume that the ball has enough speed to reach the top.

5.

A ti l ’ iti th i i i b th f ti (t2 _{4t 2) h i i}

A particle’s position on the x-axis is given by the function x=(t2_{-4t+2)m where t is in s.}

a. Make a position-versus-time graph for the interval 0 s　 　 t 　5 s. Do this by calculating and plotting x every 0.5 s from 0 s to 5 s, then drawing a smooth curve through the points

through the points.

b. Determine the particle’s velocity at t1.0 s by drawing the tangent line on your graph and measuring its slope.

c Determine the particle’s velocity at t1 0 s by evaluating the derivative at that instant c. Determine the particle s velocity at t1.0 s by evaluating the derivative at that instant.