chapter 2 motion along a straight line 2.2 motion motion: change in position in relation with an...

Chapter 2

Motion along a straight line

2.2 Motion

Motion: change in position in relation with an object of reference .

The study of motion is called kinematics.

Examples:

• The Earth orbits around the Sun

• A roadway moves with Earth’s rotation

Engineering Physics : Lecture 1, Pg 3

Position and DisplacementPosition and Displacement

Position: coordinate of position – distance to the origin (m)Position: coordinate of position – distance to the origin (m)

O

x

P

x


Example PositionExample Position

4

0 x (cm)212 1

The position of the ball is cm 2x

The + indicates the direction to the right of the origin.


Displacement Displacement ΔΔxx

Displacement: change in position x0 = original (initial) location x = final location Δx = x - x0

Example: x0 = 1m, x = 4m, Δx = 4m – 1m = 3m

1m 4m


Displacement Displacement ΔΔxx

Displacement: change in position x0 = original (initial) location x = final location Δx = x - x0

Example: x0 = 4m, x = 1 m, Δx = 1m – 4m = - 3m

1m 4m


Displacement, DistanceDisplacement, Distance

Distance traveled usually different from displacement.Distance traveled usually different from displacement. Distance always positive.Distance always positive. Previous example: always 3 m.Previous example: always 3 m.


Distance: Scalar QuantityDistance: Scalar Quantity

Distance is the path length traveled from one location to another. It will vary depending on the path.

Distance is a scalar quantity—it is described only by a magnitude.


What is the displacement in the situation depicted What is the displacement in the situation depicted bellow?bellow?

a) 3 m b) 6 m c) -3 m d) 0 ma) 3 m b) 6 m c) -3 m d) 0 m

1m 4m


What is the distance traveled in the situation depicted What is the distance traveled in the situation depicted bellow?bellow?

a) 3 m b) 6 m c) -3 m d) 0 ma) 3 m b) 6 m c) -3 m d) 0 m

1m 4m


Position, DisplacementPosition, Displacement


Motion in 1 dimensionMotion in 1 dimension In 1-D, we usually write position as x(t1 ).

Since it’s in 1-D, all we need to indicate direction is + or .

Displacement in a time t = t2 - t1 is x = x(t2) - x(t1) = x2 - x1

t

x

t1 t2

x

t

x1

x2some particle’s trajectory

in 1-D


1-D kinematics1-D kinematics

tx

tt)t(x)t(x

v12

12av

t

x

t1 t2

x

x1

x2trajectory

Velocity v is the “rate of change of position” Average velocity vav in the time t = t2 - t1 is:

t

Vav = slope of line connecting x1 and x2.


Consider limit t1 t2

Instantaneous velocity v is defined as:

1-D kinematics...1-D kinematics...

dt)t(dx

)t(v

t

x

t1 t2

x

x1

x2

t

so v(t2) = slope of line tangent to path at t2.


1-D kinematics...1-D kinematics...

tv

tt)t(v)t(v

a12

12av

Acceleration a is the “rate of change of velocity” Average acceleration aav in the time t = t2 - t1 is:

And instantaneous acceleration a is defined as:

2

2

dt)t(xd

dt)t(dv

)t(a

dt)t(dx

)t(v using


RecapRecap

If the position x is known as a function of time, then we can find both velocity v and acceleration a as a function of time!

adv

dt

d x

dt

2

2

vdx

dt

x x t ( )

x

a

vt

t

t


More 1-D kinematicsMore 1-D kinematics

We saw that v = dx / dt In “calculus” language we would write dx = v dt, which we

can integrate to obtain:

2

1

t

t12 dttvtxtx )()()(

Graphically, this is adding up lots of small rectangles:

v(t)

t

+ +...+

= displacement


High-school calculus:

Also recall that

Since a is constant, we can integrate this using the above rule to find:

Similarly, since we can integrate again to get:

1-D Motion with constant acceleration1-D Motion with constant acceleration

constt1n

1dtt 1nn

adv

dt

vdx

dt

0vatdtadtav

002

0 xtvat21

dt)vat(dtvx


RecapRecap So for constant acceleration we find:

atvv 0

200 2

1attvxx

a const

x

a

v t

t

t


Example: Displacement vs. TimeExample: Displacement vs. Time

8 m

b) What is the velocity at 5 s? Unable to determine

(no slope)c) What is the position after 10 s?

- 4 m

d) What is the velocity at 10 s?

0 m/s

e) What is the velocity during the second part of the trip?

sms

m

ss

mmvelocity /4

3

12

58

84


f) What is the velocity during the forth part of the trip?

g) What is the velocity at 15 s?

h) What is the position after 15 s?

2 m. The displacement changes 1 m every 2 seconds, so the position at 15 s is one meter morethan the position at 13 s.


Motion in One DimensionMotion in One Dimension

When throwing a ball straight up, which of the following is When throwing a ball straight up, which of the following is true about its velocity true about its velocity vv and its acceleration and its acceleration aa at the at the highest point in its path?highest point in its path?

(a)(a) BothBoth v = 0v = 0 andand a = 0a = 0..

(b)(b) v v 0 0, but , but a = 0a = 0..

(c) (c) v = 0v = 0, but , but a a 0 0..

y


Solution Solution

x

a

vt

t

t

Going up the ball has positive velocity, while coming down Going up the ball has positive velocity, while coming down it has negative velocity. At the top the velocity is it has negative velocity. At the top the velocity is momentarily zero.momentarily zero.

Since the velocity is Since the velocity is

continually changing there mustcontinually changing there must

be some acceleration.be some acceleration. In fact the acceleration is caused In fact the acceleration is caused

by gravity ( by gravity (g = 9.81 g = 9.81 m/sm/s22).). (more on gravity in a few lectures)(more on gravity in a few lectures)

The answer is (c) The answer is (c) v = 0v = 0, but , but a a 0 0. .


Galileo’s FormulaGalileo’s Formula

Plugging in for t:

atvv 0 200 at

21

tvxx

Solving for t:

avv

t 0

200

00 avv

a21

avv

vxx

)0

(220

2 xxavv


Alternate (Calculus-based) DerivationAlternate (Calculus-based) Derivation

dt

d

d

d

dt

d x

x

vva

)0

(220

2 xxavv

(chain rule)

x

vvad

d vvxa dd

v

v

x

x

x

x 000

vvxa x a ddd

)vv(2

1)-(a 2

02

0 xx

(a = constant)


Recap:Recap: For constant acceleration:

atvv 0

200 at

2

1tvxx

a const

+ Galileo and average velocity:

v)(v2

1v

)x2a(xvv

0av

02

02


Problem 1Problem 1

A car is traveling with an initial velocity v0. At t = 0, the driver puts on the brakes, which slows the car at a rate of ab

x = 0, t = 0ab

vo


Problem 1...Problem 1...

A car is traveling with an initial velocity v0. At t = 0, the driver puts on the brakes, which slows the car at a rate of ab. At what time tf does the car stop, and how much farther xf does it travel?

x = xf , t = tf

v = 0

x = 0, t = 0ab

v0



Above, we derived: v = v0 + at

Realize that a = -ab

Also realizing that v = 0 at t = tf :

find 0 = v0 - ab tf or

tf = v0 /ab



To find stopping distance we use:

In this case v = vf = 0, x0 = 0 and x = xf

fb2

0 x)a(2v

b

20

f a2v

x

)x2a(xvv 02

02



So we found that

Suppose that vo = 29 m/s Suppose also that ab = g = 9.81 m/s2

Find that tf = 3 s and xf = 43 m

b

20

fb

0f a

v

2

1x ,

a

vt


Tips:Tips:

Read !Before you start work on a problem, read the problem

statement thoroughly. Make sure you understand what information is given, what is asked for, and the meaning of all the terms used in stating the problem.

Watch your units !Always check the units of your answer, and carry the units

along with your numbers during the calculation.

Understand the limits !Many equations we use are special cases of more general

laws. Understanding how they are derived will help you recognize their limitations (for example, constant acceleration).


Problem 2Problem 2

kmkmtot ttt 32

avav v

xttvxSince

min61.0/20

22 h

hkm

kmt km

It normally takes you 10 min to travel 5 km to school. You leave class 15 min before class. Delays caused by traffic slows you down to 20 km/h for the first 2 km of the trip, will you be late to class?

usualkm v

kmt

33

min/5.0min10

5km

km

t

xv

usual

totusual

min6min/5.0

33

km

kmt km

min12min6min6 tott Not late!


34

Example: A train of mass 55,200 kg is traveling along a straight, level track at 26.8 m/s. Suddenly the engineer sees a truck stalled on the tracks 184 m ahead. If the maximum possible braking acceleration has magnitude 1.52 m/s2, can the train be stopped in time?

m 236

m/s 52.12

m/s 8.26

2

02

2

22

22

a

vx

xavv

o

o

Know: a = 1.52 m/s2, vo = 26.8 m/s, v = 0

Using the given acceleration, compute the distance traveled by the train before it comes to rest.

The train cannot be stopped in time.


A train is moving parallel and adjacent to a highway with a constant speed of 35 m/s. Initially a car is 44 m behind the train, travelingin the same direction as the train at 47 m/s, and accelerating at 3 m/s2.What is the speed of the car just as it passes the train?

Example MeetingExample Meeting

tvxx tot 2

2

1attvtvx octo 22 )/3(

2

1)/47()/35(44 tsmtsmtsmm

to = 0 s, initial timet = final timexo = 44 mvt = 35 m/s, constantvoc = 47 m/s, initial speed of carvc = final speed of car

atvv occ

2

2

1attvx occ

Car’s equation of motion

Train’s equation of motion:

To meet: xc = xt

25.112440 tt t = 2.733 s

smssmsmvc /199.55733.2)/3(/47 2

Oxo xt =

xc


Recap Recap Scope of this courseScope of this course Measurement and Units Measurement and Units (Chapter 1)(Chapter 1)

Systems of unitsConverting between systems of unitsDimensional Analysis

1-D Kinematics 1-D Kinematics (Chapter 2)Average & instantaneous velocity

and acceleration Motion with constant acceleration

Example car problemExample car problem

chapter 2 motion along a straight line 2.2 motion motion: change in position in relation with an...

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