Chapter 2 Section 2:1 Page 39
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