Chapter 2

Section 2:1Page 39

Chapter 2One Dimensional Motion

• To simplify the concept of motion, we will first consider motion that takes place in one direction.

• One example is the motion of a commuter train on a straight track.

• To measure motion, you must choose a frame of reference. A frame of reference is a system for specifying the precise location of objects in space and time.

Choose frame of reference

0 1 2 3 4 5 6 7-1-2-3-4-5-6-7-8-9

Chapter 2Displacement

x = xf – xi displacement = final position – initial position

• Displacement is a change in position.• Displacement is not always equal to the distance

traveled.• The SI unit of displacement is the meter, m.

Chapter 2Positive and Negative Displacements

Section 1 Displacement and Velocity

Chapter 2Average Velocity

• Average velocity is the total displacement divided by the time interval during which the displacement occurred.

• In SI, the unit of velocity is meters per second, abbreviated as m/s.

Average velocity =change in positionchange in time

displacement time interval=

v avg =∆x∆t =

xf –xi

tf - ti

Chapter 2Velocity and Speed

• Velocity describes motion with both a direction and a numerical value (a magnitude).

• Average speed is equal to the total distance traveled divided by the time interval.

• Speed has no direction, only magnitude.

distance traveledtime interval

average speed =

Assignments

• Practice A page 44: 1,3,4,5

• Given:• Asked:• Formula:• Substitute:• Answer with units

Interpreting Velocity Graphically

v avg =∆x∆t

xf –xi

tf - ti

=

riserun slope = =

change in vertical coordinateschange in horizontal coord

Time (s)

Disp

lace

men

t (m

)

Chapter 2

Interpreting Velocity Graphically, continued

The instantaneous

velocity at a given time can be determined by measuring the slope of the line that is tangent to that point on the position-versus-time graph.

The instantaneous velocity is the velocity of an object at some instant or at a specific point in the object’s path.

Velocity from Graph Velocity not Constant

Time (s)

Disp

lace

men

t (m

)

Assignments

• Practice A Page 44: 1,3,4,5• Section Review Page 47 Problems 1, 3, 4, 6

Acceleration and Motion • Acceleration is the change in velocity

divided by the time it takes for the change to occur.

• Acceleration has a magnitude and a direction.

• If an object speeds up, the acceleration is in the direction that the object is moving.

• If an object slows down, the acceleration is opposite to the direction that the object is moving.

Acceleration EquationsAverageAcceleration (m/s2) =

∆ v ∆t

a avg =

Change in velocity (m/s) time (s)

vf -_vi tf - ti

=

f denotes finalI denotes initial (starting)vf final velocityvi initial velocity

tf final timeti initial time∆t is always +

Practice B page 49 1-3• Given:

• Asked:• Formula:

• Substitute:• Do the math:• Answer:

∆ v ∆t

a = vf -_vi tf - ti

=

Assignment

• Practice A Page 44: 1,3.4.5• Section Review Page 47 Problems 1, 3-6• Practice B Page 49 Problems 1-3

Practice B page 49: 1

• Given:

• Asked:• Formula:

• Substitute:• Do the math:• Answer:

∆ v ∆t

a = vf -_vi tf - ti

=

Practice B page 49: 2

• Given:

• Asked:• Formula:

• Substitute:• Do the math:• Answer:

∆ v ∆t

a = vf -_vi tf - ti

=

Practice B page 49: 3

• Given:

• Asked:• Formula:

• Substitute:• Do the math:• Answer:

∆ v ∆t

a = vf -_vi tf - ti

=

AccelerationDirection and Magnitude

Page 50

Chapter 2Changes in Velocity, continued

• Consider a train moving to the right, so that the displacement and the velocity are positive.

• The slope of the velocity-time graph is the acceleration.

– When the velocity in the positive direction is increasing, the acceleration is positive, as at A.

– When the velocity is constant, there is no acceleration, as at B.

– When the velocity in the positive direction is decreasing, the acceleration is negative, as at C.

Positive AccelerationAn airliner starts at rest and then reaches a speed

(velocity) of 80 m/s in 20 s. What is its acceleration?

a = vf -_vi tf - ti

a = 80 m/s -_0 m/s 20s - 0s 80 m/s 20s

a = 4 m/s2

Negative AccelerationA skateboarder is moving in a straight line at a constant

speed of 3 m/s and comes to a stop in 2s. What is her acceleration?

a = vf -_vi tf - ti

a = 0 m/s -_3 m/s 2s - 0s - 3 m/s 2s

a =- 1.5 m/s2

Chapter 2Velocity and Acceleration (P51)

Displacement with constant acceleration

vavg = ∆x ∆t

vavg =

vf + vi

2 ∆x ∆t

vf + vi

2

∆x = ½(vf+vi ) ∆t

=

Practice C page 53, 1-4• Given:

• Asked:• Formula: ∆x = ½(vf+vi ) ∆t

• Substitute:

• Do the math:• Answer:

Assignments

• Practice A Page 44: 1,3.4.5• Section Review Page 47 Problems 1, 3-6• Practice B Page 49 Problems 1-3• Practice C page 53, 1-4

Velocity with constant accelerationFinal Velocity (p54)

∆ v ∆t

a = vf -_vi ∆t

=

a ∆t = vf - vi

vf = vi + a ∆t

Velocity with constant accelerationDisplacement from acceleration and

initial velocity (p54) ∆x = ½(vf+ vi ) ∆t derived previously

∆x = ½(vi + a ∆t + vi ) ∆t

∆x = ½(2vi + a ∆t) ∆t

∆x = vi ∆t + 1/2a ∆t2

vf = vi + a ∆t derived previously

Practice D page 55, 1-4

• Given:

• Asked:• Formula: ∆x = vi ∆t + 1/2a ∆t2

• Substitute:

• Do the math:• Answer:

vf = vi + a ∆t

Assignments

• Practice A Page 44: 1,3.4.5• Section Review Page 47 Problems 1, 3-6• Practice B Page 49 Problems 1-3• Practice C page 53, 1-4• Practice D page 55, 1-4

Final Velocity from Vi and ∆x (P56) ∆x = ½(vf+vi ) ∆t Derived Previously

2 ∆x = (vf+vi ) ∆t

= ∆t _2 ∆x_(vf+vi )

vf = vi + a ∆t Derived Previously

vf = vi + a _2 ∆x_

(vf+vi )

Derivation (cont)

vf = vi + a _2 ∆x_

(vf+vi )

vf - vi = a _2 ∆x_

(vf+vi )(vf - vi

)(vf+vi ) = a 2 ∆x

vf 2- vi

2 = 2a∆x

vf 2 = 2a ∆x +vi

2

Assignments

• Practice A Page 44: 1,3.4.5• Section Review Page 47 Problems 1, 3-6• Practice B Page 49 Problems 1-3• Practice C page 53, 1-4• Practice D page 55, 1-4 • Practice E page 57-58 (2a,b,c, 3,a,b,4,5)

Sample E page 57-58 (2-5)

• Given• Asked• Formula• Substitute• Do the math, do the math also on the units• Answer – compare units (dimensional analysis)

Schedule• Tuesday 9/4 Page 44• Wednesday 9/5 Page 47• Thursday 9/6 Page49• Friday 9/7 Page 53• Monday 9/10 Page 55• Tuesday 9/11 Page 57-58• Wednesday 9/12 Section Review P59:4-6• Thursday 9/13 Falling Objects (P64&65)• Friday 9/14 Problems P64:1,3 P65:4,5,6• Monday 9/17 Lab• Tuesday 9/18 Chapter Review• Wednesday 9/19 Chapter 2 Test

Equations for Constantly Accelerated Straight-Line Motion

Assignments

• Practice A Page 44: 1,3.4.5• Section Review Page 47 Problems 1, 3-6• Practice B Page 49 Problems 1-3• Practice C page 53, 1-4• Practice D page 55, 1-4 • Sample E page 57-58 (2-5)• Section Review P 59: 4-6

Page 59 Problem 5

Free Fall (p61)

• Free fall- a situation in which the only force on the object is the force of gravity.

• The force of gravity produces a constant acceleration.

• g• g=-9.81 m/s2

• Free fall can be up – see page 61

Assignments• Practice A Page 44: 1,3.4.5• Section Review Page 47 Problems 1, 3-6• Practice B Page 49 Problems 1-3• Practice C page 53, 1-4• Practice D page 55, 1-4 • Sample E page 57-58 (2-5)• Section Review P 59: 4-6• Sample F page 64: 1,3• Section Review page 65: 4-6• Chapter Review page 68- :3,8,17,22,31,37,46

Problem Solving

• Given:

• Asked:• Formula:• Substitute

• Answer with Units