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Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley

PowerPoint® Lectures forUniversity Physics, Twelfth Edition

– Hugh D. Young and Roger A. Freedman

Lectures by James Pazun

Chapter 2

Motion Along a Straight Line


Objectives for Chapter 2• To study motion along a straight line

• To define and differentiate average and instantaneous linear velocity

• To define and differentiate average and instantaneous linear acceleration

• To explore applications of straight-line motion with constant acceleration

• To examine freely falling bodies

• To consider straight-line motion with varying acceleration


Homework

Read 52-66, gain full understanding of the figures and memorize the equations.

2.5, 2.9, 2.19, 2.31, 2.35, 2.39, 2.55, 2.75(Bonus)


Introduction

• Runners accelerate out of starting blocks, run faster, then slow down as the race ends. We will be able to describe this motion.

• A tree drops in front of my car. Can I stop in time? We will be able to answer this question.

• This study of motion … kinematics … is common to our lives yet full of interesting features. We begin with mechanics, the study of the relationships among force, matter, and motion. Later we will cover dynamics, the relation of motion to its causes.


Displacement, time, and the average velocity—Figure 2.1To analyze an object’s motion, we need a common frame of reference.The object moves from one position to another. This change in position,

, is the final position minus the initial position, as shown below.The dragsters average velocity is defined as the displacement divided by the time interval.

xΔ


The truck, its motion, and a graph—Figures 2.2 and 2.3


Average and instantaneous velocitiesRaces, in general, are tests to see who has the highest average velocity: the same displacement but shortest time interval yields the largest Vav.To describe motion in greater detail, we need to describe velocity at any specific point in the objects path. This is called instantaneous velocity.To get this velocity, lets consider average velocity:

The instantaneous velocity is the limit of the average velocity as the time interval approaches zero; it equals the instantaneous rate of change of position with time.

av xxvt−

Δ=Δ

0limx t

x dxvt dtΔ →

Δ= =

Δ


ExampleA cheetah is crouched in ambush 20 m to the east of an observer’s blind. At time t = 0 the cheetah charges an antelope in a straight line. Later analysis of a videotape shows that during the first 2.0 s of the attack, the cheetah’s coordinate x varies with time according to the equation x = 20 m + (5.0 m/s2)t2.

a) Find the displacement of the cheetah during the interval between t1 = 1.0 s and t2 = 2.0 s.

b) Find the average velocity during the same time interval.

c) Find the instantaneous velocity at time t1 = 1.0 s by taking Δt = 0.1 s, then Δt = 0.01 s, then Δt = 0.001 s.

d) Derive a general expression for the instantaneous velocity as a function of time, and from it find vx at t = 1.0 s and t = 2.0 s.


A safari and a chase—Figure 2.6


Follow the motion of a particle—Figure 2.8

Look at figure 2.9 on page 47. Describe the motion at:P?Q?R?S?


Average and Instantaneous AccelerationJust as velocity describes the rate of change of position with time, acceleration describes the rate of change of velocity with time.

We define the average acceleration aav-x of the particle as it moves from P1 to P2 to be a vector quantity whose x-component is ΔVx, the change in the x-component of velocity, divided by the time interval Δ t.Instantaneous acceleration is found by bringing our two points (like in figure 2.11 on page 49) closer and closer together. We do this until the time interval approaches zero. That leaves us with . . .A limit!!

The instantaneous acceleration is the limit of the average acceleration as the time interval approaches zero.

2 1

2 1

x x xav x

v v vat t t−

− Δ= =

− Δ

2

20lim x x

x t

v dv d dx d xat dt dt dt dtΔ →

Δ ⎛ ⎞= = = =⎜ ⎟Δ ⎝ ⎠


The average acceleration—Example 2.2


Example—Instantaneous AccelerationSuppose the velocity vx of the car in figure 2.11 at any time t is given by

the equation:

a) Find the change in velocity of the car in the time interval between t1= 1.0 s and t2 = 3.0 s

b) Find the average acceleration in this time interval.

c) Find the instantaneous acceleration at time t1 = 1.0 s by taking Δt = 0.1 s, Δt = 0.01 s, and Δt = 0.001 s.

d) Derive an expression for the instantaneous acceleration at any time, and use it to find the acceleration at t = 1.0 s and t = 3.0 s.

( )3260 0.50m m

sx sv t= +


Finding the acceleration—Figures 2.12, 2.13 & 2.14


Motion with Constant AccelerationA body in free fall, an object sliding down an incline plane or across a rough surface can all have a constant acceleration, where the velocity changes at a constant rate.For a constant acceleration, the average acceleration is the same as the instantaneous acceleration.

Giving (with t1 = 0):

2 1

2 1

x xx

v vat t−

=−

0x x xv v a t= +


Motion with constant acceleration—Figures 2.16 and 2.17


The equations of motion under constant acceleration• Pages 53 to 55 follow the

derivation of four equations of constant acceleration. They are shown at right.

• Special mention is made of these four equations because they will permeate our study of kinematics (linear and circular, too).

• Follow the steps in Problem-Solving Strategy 2.1 for any problem involving motion with constant acceleration.

vx = vox + axt

x = xo + voxt + 1/2axt2

vx2 = vox

2 + 2ax(x − xo)

(x − xo) = {(1/2)(vox + vx)}t


Use the equations to study motorcycle motionA motorcyclist heading east though a small Iowa city accelerates after he passes the signpost marking the city limits. His acceleration is a constant 4.0 m/s2. At time t = 0 he is 5.0m east of the signpost, moving east at 15 m/s.

Find his position and velocity at time t = 2.0 s.

Where is the motorcycle when his velocity is 25 m/s?


Study two bodies with different accelerations A motorist traveling with a constant velocity of 15 m/s passes aschool crossing corner, where the speed limit is 10 m/s. Just as the motorist passes, a police officer on a motorcycle stopped atthe corner starts off in pursuit with constant acceleration of 3.0 m/s2. How much time elapses before the officer catches up with the motorist?What is the officer’s speed at that point?What is the total distance each vehicle has traveled at that point?


Free fall—Figure 2.22

• A strobe light begins to fire as the ball is dropped.

• Notice how the space between images increases as the ball’s velocity grows.

• Ball accelerates at a constant rate: g = -9.81 m/s2


Free fall• Aristotle thought that

heavier bodies would fall faster. Galileo is said to have dropped two objects, one light and one heavy, from the top of the Leaning Tower of Pisa to test his assertion that all bodies fall at the same rate.

• Astronaut Dave Scott tested this himself by dropping a hammer and a feather on the moon.


A Freely-Falling CoinA one-euro coin is dropped from the Leaning Tower of Pisa. It starts from rest and falls freely. Compute its position and velocity after 1.0 s, 2.0 s and 3.0 s.


Free fall III—Figure 2.24You throw a ball vertically upward from the roof of a tall building. The ball leaves your hand at a point even with the roof railing with an upward speed of 15.0 m/s; the ball is then in free fall. On its way back down, it just misses the railing. At the location of the building, g = 9.80 m/s2. Find:

The position and velocity of the ball 1.00 s and 4.00 s after leaving your hand.

The velocity when the ball is 5.00 m above the railing.

The maximum height reached and the time at which it is reached.

The acceleration of the ball when it is at its maximum height.


Is velocity zero at the highest point?—Figure 2.25

Find the time when the ball is 5.00 m below the roof railing.


When acceleration is not constant—Figures 2.26 and 2.28

• Even the smoothest sports car does not move with constant acceleration.

• The motion may be integrated over many small time windows.


Analysis of motion—Figure 2.29

• All components of motion under changing acceleration may be examined.

• Refer to Examples 2.9 and 2.10.

motion along a straight line - fcps 2 power...a cheetah is crouched in ambush 20 m to the east of an...

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