Ch 2. Motion ina Straight Line

Definitions1. Kinematics - Motion

Kinetic Energy - Energy associated with motion

2. Motion in physics is broken down into categoriesa.) Translational Motion - motion such that an object moves from one

position to another along a straight line.b.) Rotational Motion - motion such that an object moves from one

position to another along a circular path.c.) Vibrational Motion - motion such that an object moves back and

forth in some type of periodicity.

Example: Diatomic Molecule Moving Through Space.

Rotational

Vibrational

Translational

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“Dinophysics : Velocity-Raptor”

Speed1. Speed - How fast an object is moving regardless of what direction it is

moving.

Example 1. Traveling from your parking space at Conestoga to New York City and back to Conestoga. Find your avg. speed.

x(mi)

y(mi)

up

back

NY

Conestoga

Average Velocity and DisplacementDisplacement - Change in position (straight line distance with direction)

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One way travel = 130 mi.Total Distance Traveled = 260 mi.Total time elapsed = 5.2 hrs. or (5 hrs 12 min)

Must specify a coordinate system.

Example: Cartesian coordinate system

x = Change in x = xf xi m- 2 m = + 4 m x 6

Avg. Velocity - How fast an object is moving and in what direction it is moving.

Average Velocity = Change in positionChange in time

Average Velocity = xt

x xt tf if i

x(m)

y(m)

xi= 2m

x1

xf= 6m

x2

xi = xinitial= Initial Position

xf = xfinal = Final Position

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Gives the direction“Delta”

Delta x is the displacement

Notation for Displacement & Vel. Example Problem

x = x “hat”, and has a value of one. The sole purpose of x is to indicate the direction

Example Problem:

A particle initially at position x = 5 m at time t= 2 s moves to position x = -2 m and arrives at time t = 4 s.

a.) Find the displacement of the particle.

b.) Find the average speed and velocity of the particle.

x(m)

y(m)

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Example Problem 1 revisited

Example 1. Traveling from your parking space at Conestoga to New York City and back to Conestoga. The straight line distance from Conestoga to Y is 97 mi.

x(mi)

y(mi)

up

back

NY

Conestoga 970

a.) What was the avg. speed from Conestoga to NY?b.) What was the avg. velocity from Conestoga to NY?c.) What was the avg. speed for the round trip?d.) What was the avg. velocity for the round trip?

Note: Speed in PATH DEPENDENTVelocity is PATH INDEPENDENT. It only depends on the initial and final positions.

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One way travel = 130 mi.Total Distance Traveled = 260 mi.Travel time Con. to NY = 2.6 hrs.Travel time NY to Con. = 2.6 hrs.

Scalar vs. Vector Quantities

Scalar - Quantity that has magnitude only.

- Mass - Speed

- Length - Energy

Vector - A quantity that has both magnitude and direction.

- Position - Acceleration

- Velocity - Forces

Example: Length vs. Position

x(m)21 3 5 640-2 -1-3-4-5-6

Pt. APt. B

As measured from the origin the length to pt. A is 3m.

To specify the Pt. A in space you must reference it to the origin and then Pt. A = +(3 m) x

or, the Position Vector A = +(3 m) x

x = x “hat”, and is called a unit vector in the x-direction. It has a

magnitude of one (hence the name unit) and is used solely to

specify direction.

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Position vs Time Graph

x Y1 Y2

Time (s) Position (m) Position (m)0 0 01 1 52 4 103 9 154 16 205 25 25

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Movement 1

Movement 2

A

B C

D

E

F

-150

-100

-50

0

50

100

150

200

250

300

0 10 20 30 40 50 60Time (s)

Posi

tion

(m)

Position vs. Time Graph for a Complete Trip

x yTime (s) Position (m)

0 010 20020 20025 15045 -10060 0

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Find the average velocity as the object moves from:a.) A to B b.) B to C c.) C to D.d.) A to E

Velocity vs. Time (Constant Velocity)

Velocity vs. Time Graph for a Complete Trip

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Velocity Function

0

1

2

3

4

0 1 2 3 4 5t(sec)

v (m

/s)

Position Function

0

1

2

3

4

5

6

7

8

0 1 2 3 4 5

t(sec)

x(m

)

x

t

v x

tave

Area!

x

A

B C

D

E

F

-150

-100

-50

0

50

100

150

200

250

300

0 10 20 30 40 50 60Time (s)

Posi

tion

(m)

A B

B C

CD

D E

E F

-25

-20

-15

-10

-5

0

5

10

15

20

25

0 10 20 30 40 50 60

Time (s)

Velo

city

(m/s

)

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Instantaneous Velocity

Recall: (Average velocity)

Consider the function x(t): x t m + 10 m

s t - 0.5 m

s t

22

44

3

A. 2 s < t < 3.5 s t = 1.5 sec

B. 2 s < t < 2.2 s t = 0.2 sec

0

10

20

30

40

50

60

0 1 2 3 4t(sec)

x(m

)

0

10

20

30

40

50

60

0 1 2 3 4t(sec)

x(m

)

A. B.

t (sec) vave (m/s)1 17.5

0.2 23.450.01 23.980.001 23.998

Define: The instantaneous velocity at time ti is the slope of the line tangent to the curve X(t) at the time ti.

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slope = 23.5 m/sslope = 10.3m/s

The instantaneous velocity at the time t = ti is the limiting value we get by letting the upper value of the tf approach ti. Mathematically this is expressed as:

v t

dX tdt lim

X t X tt tt f ti

f i

f i

The velocity function v t is the time derivative of the position function X t .

Differentiation (Calculus)

Acceleration

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When the instantaneous velocity of a particle is changing with time, the particle is accelerating

(Average Acceleration)

Units:

Example: If a particle is moving with a velocity in the x-direction given by

a.) What is the average acceleration over the time interval 6 t 12 s s

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Positive and Negative Accelerations

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Example: Instantaneous Acceleration

a.) Find aavg. over the time interval 5 t 8

3 t 6

b.) What is the acceleration at time t = 6 s ?c.) What is the acceleration when the velocity of the particle is zero?

Velocity vs. Time

-15

-10

-5

0

5

10

15

20

25

30

35

0 2 4 6 8 10

time (s)

velo

city

(m/s

)

time (s) vel. (m/s)0 -101 -22 -53 54 125 146 127 218 30

v(m/s)

t (s)

A

C

E

D

B

F

A→B:

B→C:

C→D:

D→E:

E→F:

Special Case: Constant Acceleration

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a(m/s2)

t (s)ti = 0 tf = t

a

0

We make the assumption that the acceleration does not change. Near the surface of the earth, (where most of us spend most of our time) the acceleration due to gravity is approximately constant ag = 9.8 m/s2

Area! Slope!

v(m/s)

t (s)ti = 0 tf = t

vf

0

x(m)

t (s)ti = 0 tf = t

xf

xi

Area! Slope!

vi

v = v + a tf i

x x + v tf i i2= + 1

2a t

1.

2.

Solving for the 3rd constant acceleration equation

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Solve equation 1 for t and substitute t into equation 2 to get the following equation.

FREE-FALL ACCELERATION(9.8 m/s2 = 32 ft/s2)

Consider a ball is thrown straight up.It is in “Free Fall” the moment it leaves you hand.

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3.

Plot y(t) vs. t for the example above.

Plot v(t) vs. t

FINAL NOTES ON CH 2.

Remember , when going between the following graphs

tf t/2

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y(t)

tf t/2

Problem Solving with the constant acceleration equations

1. Write down all three equations in the margin2. a = 9.8 m/s2 for free fall problems3. Analyze the problem in terms of initial and final sections.

CH 2 Describing Motion: Kinematics in 1-DPractice Questions

_______1. A car starts from Hither, goes 50 km in a straight line to Yon, immediately turns around, and returns to Hither. The for this round trip 2 hours. The magnitude of the average velocity of the car for this round trip is:

A. 0B. 50 km/hrC. 100 km/hr

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x(t)

Area Under Curve

Slopev(t)

a(t)

x

t

A

x

t

B

x

t

C

x

t

D

x

t

E

D. 200 km/hrE. Cannot be calculated without knowing the acceleration

_______2. An object starts from rest at the origin and moves along the x axis with a constant acceleration of 4 m/s2. Its average velocity as it goes from x = 2 m to x = 8m is:

A. 1 m/sB. 2 m/sC. 3 m/sD. 5 m/sE. 6 m/s

_______3. A ball is in free fall. Its acceleration is:A. Downward during both ascent and descentB. Downward during ascent and upward during descentC. Upward during ascent and downward during descentD. Upward during both ascent and descentE. Downward at all times except at the very top, when it is zero

_______4. The coordinate-time graph of an object is a straight line with a positive slope. The object has:

A. Constant displacementB. Steadily increasing accelerationC. Steadily decreasing accelerationD. Constant velocityE. Steadily increasing velocity

_______5. A car accelerates from rest on a straight road. A short time later, the car decelerates to a stop and then returns to its original position in a similar manner. Which of the following graphs best describes the motion?

Answers: 1.A 2.E 3.A 4.D 5.E

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