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MotionMotion
PSc.1.1 OBJECTIVE: Understand motion in terms of speed, velocity,
acceleration, and momentum.
ObjectivesObjectives
PSc.1.1.1 Interpret all motion as relative
to a selected reference point. Identify distance and
displacement as a scalar-vector pair.
ObjectivesObjectives
PSc.1.1.1, cont.Describe motion qualitatively
and quantitatively in terms of
•an object’s change of position,
•distance traveled,
•and displacement.
Reference PointReference Point
For all motion problems we need a reference point... a non-moving point from which
motion is measured.
Reference PointReference Point
A reference point is a point or object that is used to measure what the distance and direction to another object is.
In the picture above, the reference point is the yard line.
Reference PointReference Point
In the picture above your house is used as a reference point to determine where your friend’s house is.
Reference PointReference Point
MotionMotion
An object is in motion when the distance between the object and the reference point is changing.
Reference point
Motion
MotionMotion
The person is in motion when the fence is used as a reference point because the distance between the person and the fence is changing.
MotionMotion
The person is NOT in motion when the ground is used as a reference point because the distance between the person and the ground is not changing.
DistanceDistanceDistance is the space traveled
between the reference point and the object or ending point.
The distance between the reference point and the blue football player is 3 m (meters).
DistanceDistance• Problem #1:
Suppose a runner jogs to the 50-m North mark and then turns around and runs back to the 20-m mark South. Determine her distance.
The runner travels 50 m in the original direction (north) plus 30 m in the opposite direction (south), so the total distance she ran is 80 m.
PositionPosition
Position is an object’s location compared to a reference point.
It includes both a distance and direction.
PositionPosition
If City Hall is the reference point, the position of the library is 500 m East.
DisplacementDisplacement
Displacement is a change of position in a certain direction, not the total distance traveled.
DisplacementDisplacement
The displacement is the shorter directed distance from start to stop (green arrow).
start
stop
DisplacementDisplacement
If two displacement vectors are going in the same direction add to find the total displacement.
If two displacement vectors are going in opposite directions subtract to find the total displacement.
DisplacementDisplacement
Problem #2:
A man walks 54.5 meters east and then and an additional 30.0 meters east. Calculate his displacement relative to where he started.
54.5 m, E 30.0 m, E
84.5 m, E
DisplacementDisplacement
Problem #3:
A man walks 54.5 meters east and then 30.0 meters west. Calculate his displacement relative to where he started.
54.5 m, E
30.0 m, W
24.5 m, E
Distance vs DisplacementDistance vs Displacement
Problem #4: Suppose a bus starts from terminal A, travels 1500 m to reach terminal B and then returns to terminal A. What is a) the distance traveled and
3000 m
b) the displacement traveled?
0 m
Vectors and ScalarsVectors and Scalars
A scalar quantity is any quantity that has a magnitude, but NO direction associated with it.
Scalar Example
Magnitude
Speed 20 m/s
Distance 10 m
Age 15 years
Heat 1000 calories
Magnitude – A numerical value with units.
Vectors and ScalarsVectors and Scalars
Distance is a scalar quantity.
The distance between the two books is 15 units.
Vectors and ScalarsVectors and Scalars
The distance points A and B is 6 units.
Vectors and ScalarsVectors and ScalarsA vector is represented by an arrow. A vector gives 2 pieces of
information: magnitude and direction.
42°
5 m/s
The magnitude is 5 m/s.
The direction is 42°.
Vectors and ScalarsVectors and Scalars
The length of the arrow represents the magnitude (how far, how fast, how strong, etc, depending on the type of vector).
The arrow points in the direction of the force, motion, displacement, etc. It is often specified by an angle.
Vectors and ScalarsVectors and Scalars
Displacement is a vector quantity.
The displacement between points A and B found by using the Pythagorean Theorem.
Vectors and ScalarsVectors and Scalars
(5)2 + (1)2 = (AB)2
The x-displacement is 5 units.The y-displacement is 1 unit.
25 + 1 = (AB)2
√ 26 = AB
5.1 units = AB
Vectors and ScalarsVectors and Scalars
The direction is northeast.
Vectors and ScalarsVectors and Scalars
Distance and displacement are referred to as a scalar-vector pair.
Vectors and ScalarsVectors and Scalars
Problem #5:
John runs 3 km north, then turns and walks 4 km south.
a) Calculate the total distance covered.
b) Calculate the total displacement.
7 km
1 km south
Vectors and ScalarsVectors and Scalars
Problem #6:
During a ride in a hot air balloon, a group of people are carried 50 km [North], 65 km [West] and then 75 km [South].
a) Calculate the total distance covered.190 km
b) Calculate the total displacement.70 km southwest
ObjectivesObjectives
PSc.1.1.2 Compare speed and velocity as a scalar-
vector pair. Velocity is a relationship between
displacement and time. Apply concepts of average speed and
average velocity to solve conceptual and quantitative problems.
Explain acceleration as a relationship between velocity and time.
• Distance and time are important when considering motion. In order to win a race, you must cover the distance in the shortest amount of time.
• How would you describe the motion of the runners in the race?
MotionMotion
SpeedSpeed
Speed is the measurement of the change in distance for a given period of time.
The car traveled 40 m in 6 s.
SpeedSpeed
Instantaneous speed is speed at a given instant.
The speed shown on a speedometer is the instantaneous speed.
SpeedSpeed
When something is speeding up or slowing down, its instantaneous speed is changing.
Constant speed means the speed stays the same.
If an object is moving with constant speed, the instantaneous speed doesn't change.
SpeedSpeed
Average speed is total distance divided by total time.
time total
distance totalspeed avg.
SpeedSpeed
time
distancespeed
vd
t
VelocityVelocity
Velocity is the measurement of the change in displacement for a given period of time.
VelocityVelocity
Velocity is speed in a given direction and can change even when the speed is constant!
Speed and VelocitySpeed and Velocity
Speed and velocity are a scalar-vector pair.
Speed is a scalar quantity that refers to "how fast an object is moving."
Velocity is a vector quantity that refers to "the rate at which an object changes its position."
Speed and VelocitySpeed and Velocity
Since a vector quantity has a direction associated with it, velocity values can be negative if they are directed west or south.
Speed cannot be negative.
Speed and VelocitySpeed and Velocity
To calculate speed/velocity you need the distance/displacement traveled and the time the object traveled.
t
dv
A car traveled 40 m in 6 s. Determine the speed.
The speed of the car is 6.7 m/s.
t
dv
s
mv
6
40
Speed/Velocity Speed/Velocity Problem #7Problem #7
Problem Distance Displacement
1 135 miles 85 miles
2 6 m 0 m
3a 7 km 5 km NE
3b 5 blocks 3.6 blocks SE
3c 800 m 0 m
3d 5 blocks 3.6 blocks NW
3e 300 m 0 m
4 2750 km 750 km
5 180 km 170 km N
6 390 m 30 m E
7a 30 m 0 m
7b 45 m 45 m E
8 9 km 6.7 km NE
9 17 km 11.7 km SW
10 12 m 8.9 m NE
11 12 km 2 km N
Speed/Velocity Speed/Velocity Problem #8Problem #8
Your neighbor skates at a speed of 4 m/s. You can skate 100 m in 20 s. Who skates faster?
GIVEN:
Δd = 100 m
Δt = 20 s
v = ?
WORK:
v = Δd ÷ Δt
v = (100 m) ÷ (20 s)
v = 5 m/s
You skate faster!vd
t
A baseball pitcher throws a fastball at 42 m/s. If the batter is 18 m from the pitcher, how much time does it take for the ball to reach the batter?
GIVEN:
Δd = 18 m
Δt = ?
v = 42 m/s
WORK:
Δt = Δd ÷ v
Δt = (18 m) ÷ (42 m/s)
Δt = 0.43 s
vd
t
Speed/Velocity Speed/Velocity Problem #9Problem #9
A toy car moves at a constant velocity of 1.6 m/s. If it continues, how far will the car travel in 3.0 s?
GIVEN:
Δd = ?
Δt = 3.0
v = 1.6 m/s
WORK:
Δd = v • Δt
Δd = (1.6 m/s) • (3.0 s)
Δd = 4.8 m
vd
t
Speed/Velocity Speed/Velocity Problem #10Problem #10
Speed/Velocity Speed/Velocity Problem #11Problem #11
You travel 35 km in 0.4 h, followed by 53 km in 0.6 h. What is your average speed?
GIVEN:
Δd = (35 + 53) km
Δt = (0.4 + 0.6) h
v = ?
WORK:
v = Δd ÷ Δt
v = (88 km) ÷ (1.0 h)
v = 88 km/h
vd
t
Speed and VelocitySpeed and Velocity
Problem: A storm is 10 km away and is
moving at a speed of 60 km/h. Should you be worried?
It depends on the storm’s direction!
Speed and Velocity Speed and Velocity Problem Answer
1 62.2 mi/hr
2 4.76 mi/hr
3 4400 mi
4 238.2 mi
5 800 mi
6 381.5 mi
7 68.7 mi/hr
8 4.23 hr
AccelerationAcceleration
Acceleration is the rate of change of velocity (change in speed or direction).
Acceleration occurs whenever there is a change in speed, direction or both.
Acceleration occurs whenever there is a change in speed, direction, or both.
AccelerationAcceleration
AccelerationAcceleration
Positive acceleration “speeding up”
Negative acceleration “slowing down”
Signs for Velocity and Signs for Velocity and AccelerationAcceleration
AccelerationAcceleration
t
vva if
a: acceleration
vf: final velocity
vi: initial velocity
t: time
a
vf - vi
t
Acceleration Acceleration Problem #12Problem #12 A roller coaster starts down a hill at
10 m/s. 3.0 seconds later, its speed is 32 m/s. What is its acceleration?
GIVEN:
vi = 10 m/s
t = 3 s
vf = 32 m/s
a = ?
WORK:
a = (vf - vi) ÷ t
a = (32m/s - 10m/s) ÷ (3s)
a = 22 m/s ÷ 3 s
a = 7.3 m/s2a
vf - vi
t
Sound travels 330 m/s. If lightning strikes the ground 1 km away from you, how long will it take for you to hear it?
GIVEN:
v = 330 m/s
d = 1km = 1000m
t = ?
WORK:
t = d ÷ v
t = (1000 m) ÷ (330 m/s)
t = 3.03 s
vd
t
Acceleration Acceleration Problem #13Problem #13
How long will it take a car traveling 30 m/s to come to a stop if its acceleration is - 3 m/s2?
GIVEN:
t = ?
vi = 30 m/s
vf = 0 m/s
a = -3 m/s2
WORK:
t = (vf - vi) ÷ a
t = (0m/s-30m/s)÷(-3m/s2)
t = -30 m/s ÷ -3m/s2
t = 10 sa
vf - vi
t
Acceleration Acceleration Problem #14Problem #14
Speed, Velocity and Accel.Speed, Velocity and Accel.
Problem Answer
1 75.9 km/hr
2 -2 m/s2
3 30 km
4 5 m/s2
5 -3.33 m/s2
6 24 m/s
7 0.1 s
8 8.6 m/s
ObjectivesObjectives
PSc.1.1.2, cont. Using graphical analysis, solve for
displacement, time, and average velocity. Analyze conceptual trends in the displacement vs. time graphs such as constant velocity and acceleration.
ObjectivesObjectives
PSc.1.1.2, cont. Using graphical analysis, solve for
velocity, time, and average acceleration. Analyze conceptual trends in the velocity vs. time graphs such as constant velocity and acceleration.
• The independent variable is the variable that the experimenter manipulates or changes, on purpose.
• The dependent variable is a variable that changes depending on some other factors.
GraphingGraphing
Put the dependent variable on the ‘y-axis’
and the independent variable on the ‘x-axis.’
GraphingGraphing
Slope of a graphSlope of a graph
The slope of a graph is equal to the ratio of rise to run.
• The motion of an object over a period of time can be shown on a distance-time graph.
Graphing Motion
• Time is plotted along the horizontal axis of the graph and the distance traveled is plotted along the vertical axis of the graph.
• Each axis must have a scale that covers the range of number to be plotted.
• Once the scales for each axis are in place, the data points can be plotted.
• After plotting the data points, draw a line connecting the points.
Graphing Motion
Distance-Time GraphDistance-Time Graph
slope =
steeper slope =
straight line =
flat line =
Distance-Time Graph
A
B
faster speed
constant speed
no motion (at rest)
speed
Curve =acceleration
Distance-Time Graph
Distance-Time GraphDistance-Time Graph
0
100
200
300
400
0 5 10 15 20
Time (s)
Dis
tan
ce (
m)
Distance-Time Graph
Acceleration is indicated by a curve on a distance-time graph.
Changing slope = changing velocity
Distance-Time GraphDistance-Time Graph
What would these look like on a distance-time graph?
1. stopped
2. slow
3. fast
4. accelerating
Distance-Time GraphDistance-Time Graph
A or B: Who started out faster? A (steeper
slope)A or B: Who had a
constant speed? A (continuous
straight line)
Distance-Time Graph
A
B
Distance-Time GraphDistance-Time Graph
Describe B’s motion from 10-20 min. B stopped moving
Find their average speeds. A = (2400 m) ÷ (30 min) A = 80 m/min B = (1200 m) ÷ (30 min) B = 40 m/min
Distance-Time Graph
A
B
Distance-Time GraphDistance-Time Graph
Speed-Time GraphsSpeed-Time Graphs
0
1
2
3
0 2 4 6 8 10
Time (s)
Sp
ee
d (
m/s
)
Speed-Time Graph
slope =acceleration + slope = speeds
up negative slope =
slows down
0
1
2
3
0 2 4 6 8 10
Time (s)
Sp
ee
d (
m/s
)
Speed-Time Graph
straight line =
flat line =
constant accel.
no accel. (constant velocity)
Speed-Time GraphsSpeed-Time Graphs
0
1
2
3
0 2 4 6 8 10
Time (s)
Sp
ee
d (
m/s
)
Speed-Time Graph
Specify the time period when the object was...
slowing down 5 to 10 seconds
speeding up 0 to 3 seconds
Speed-Time GraphsSpeed-Time Graphs
Graphing MotionGraphing Motion
0
1
2
3
0 2 4 6 8 10
Time (s)
Sp
ee
d (
m/s
)
Speed-Time Graph
Specify the time period when the object was...
moving at a constant speed 3 to 5 seconds
not moving 0 & 10 seconds
ObjectivesObjectives
PSc.1.1.2, cont. Infer how momentum is a relationship
between mass and velocity of an object.
Explain change in momentum in terms of the magnitude of the applied force and the time interval that the force is applied to the object.
MomentumMomentum
Momentum refers to inertia in motion.
Momentum is a measure of how difficult it is to stop an object or a measure of “how much motion” an object has.
MomentumMomentum
Momentum quantity of motion
p = mvp: momentum (kg ·m/s)m: mass (kg)v: velocity (m/s)m
p
v
Momentum Momentum Problem #15Problem #15
Find the momentum of a bumper car if it has a total mass of 280 kg and a velocity of 3.2 m/s.
GIVEN:
p = ?
m = 280 kg
v = 3.2 m/s
WORK:
p = mv
p = (280 kg)(3.2 m/s)
p = 896 kg·m/s
m
p
v
Momentum Momentum Problem #16Problem #16
The momentum of a second bumper car is 675 kg·m/s. What is its velocity if its total mass is 300 kg?
GIVEN:
p = 675 kg·m/s
m = 300 kg
v = ?
WORK:
v = p ÷ m
v = (675 kg·m/s)÷(300 kg)
v = 2.25 m/s
m
p
v
• Sometimes the momentum of an object changes.
• Change in momentum can be called impulse.
Change in MomentumChange in Momentum
Force applied on everyday objects results in a change in velocity.
Change in MomentumChange in Momentum
• When playing with a paddleball, the less time the ball is in contact with the paddle, the more force applied to the ball.
Change in MomentumChange in Momentum
• If time is extended, less force will be applied to the object.
Change in MomentumChange in Momentum
The purpose of seat belts, air bags, and padded dashboards is to extend the time during which you come to rest during a crash. These safety devices help reduce the forces exerted on you.
Change in MomentumChange in Momentum
A 1000 kg car moving at 30 m/s (p = 30,000 kg m/s) can be stopped by 30,000 N of force acting for 1.0 s (a crash) OR
Change in MomentumChange in Momentum
by 3000 N of force acting for by 3000 N of force acting for 10.0 s (normal stop).10.0 s (normal stop).
Change in MomentumChange in Momentum
The same change in momentum can be accomplished by a small force acting for a long time or by a large force acting for a short time.
Change in MomentumChange in Momentum
F·t = m·v
where v = vf - vi
Use the diagrams below to determine the magnitude of changes in velocity of the same ball.
∆v = 15 m/s
∆v = 58 m/s
Change in MomentumChange in Momentum
Calvin throws a 0.450 kg snowball at Susie. Calculate the initial speed of the snowball if its change in momentum is 3.40 kg.m/s.
(7.56 m/s)
Change in Momentum Change in Momentum Problem Problem #17#17
A 0.40 kg soccer ball approaches Joe horizontally with a velocity of 18 m/s north. He strikes the ball and causes it to move in the opposite direction with a velocity of 25 m/s. Calculate the magnitude of the change in momentum of the ball.
(17 kg.m/s)
Change in Momentum Change in Momentum Problem Problem #18#18
Calvin strikes a 0.0450 gram golf ball with a club. The force applied to the ball is 1900. N. If the club is in contact with the ball for 0.00500 s, what is the change in velocity of the golf ball?
(211 m/s)
Change in Momentum Change in Momentum Problem Problem #19#19
• If a pitcher pitches the ball at a speed of 40.2 m/s, and Calvin sends it back with a speed of 49.1 m/s, determine the time the ball is in contact with the bat. Calvin applies a force of 12,000 N to the 0.142 kg baseball.
(0.00106 s)
Change in MomentumChange in MomentumProblem #20Problem #20
Conservation of MomentumConservation of Momentum
Law of Conservation of Momentum The total momentum in a group
of objects doesn’t change unless outside forces act on the objects.
pbefore = pafter
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