Motion PSc.1.1 OBJECTIVE: Understand motion in terms of speed, velocity, acceleration, and momentum.

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.

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.

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.

In the picture above your house is used as a reference point to determine where your friend’s house is.

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.

The displacement is the shorter directed distance from start to stop (green arrow).

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.

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

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?

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.

Distance is a scalar quantity.

The distance between the two books is 15 units.

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.

The magnitude is 5 m/s.

The direction is 42°.

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.

Displacement is a vector quantity.

The displacement between points A and B found by using the Pythagorean Theorem.

(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

The direction is northeast.

Distance and displacement are referred to as a scalar-vector pair.

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.

1 km south

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

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

distancespeed

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."

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.

To calculate speed/velocity you need the distance/displacement traveled and the time the object traveled.

A car traveled 40 m in 6 s. Determine the speed.

The speed of the car is 6.7 m/s.

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

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 = Δd ÷ Δt

v = (100 m) ÷ (20 s)

v = 5 m/s

You skate faster!vd

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

Δt = Δd ÷ v

Δt = (18 m) ÷ (42 m/s)

Δt = 0.43 s

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

Δd = v • Δt

Δd = (1.6 m/s) • (3.0 s)

Δd = 4.8 m

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 = Δd ÷ Δt

v = (88 km) ÷ (1.0 h)

v = 88 km/h

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.

Positive acceleration “speeding up”

Negative acceleration “slowing down”

Signs for Velocity and Signs for Velocity and AccelerationAcceleration

vva if

a: acceleration

vf: final velocity

vi: initial velocity

t: time

vf - vi

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 = (vf - vi) ÷ t

a = (32m/s - 10m/s) ÷ (3s)

a = 22 m/s ÷ 3 s

a = 7.3 m/s2a

vf - vi

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 = d ÷ v

t = (1000 m) ÷ (330 m/s)

t = 3.03 s

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:

vi = 30 m/s

vf = 0 m/s

a = -3 m/s2

t = (vf - vi) ÷ a

t = (0m/s-30m/s)÷(-3m/s2)

t = -30 m/s ÷ -3m/s2

t = 10 sa

vf - vi

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

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.

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

faster speed

constant speed

no motion (at rest)

Curve =acceleration

Distance-Time Graph

0 5 10 15 20

Time (s)

Distance-Time Graph

Acceleration is indicated by a curve on a distance-time graph.

Changing slope = changing velocity

What would these look like on a distance-time graph?

1. stopped

2. slow

3. fast

4. accelerating

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

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

Speed-Time GraphsSpeed-Time Graphs

0 2 4 6 8 10

Time (s)

Speed-Time Graph

slope =acceleration + slope = speeds

up negative slope =

slows down

0 2 4 6 8 10

Time (s)

Speed-Time Graph

straight line =

flat line =

constant accel.

no accel. (constant velocity)

0 2 4 6 8 10

Time (s)

Speed-Time Graph

Specify the time period when the object was...

slowing down 5 to 10 seconds

speeding up 0 to 3 seconds

Graphing MotionGraphing Motion

0 2 4 6 8 10

Time (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

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

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:

m = 280 kg

v = 3.2 m/s

p = mv

p = (280 kg)(3.2 m/s)

p = 896 kg·m/s

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 = p ÷ m

v = (675 kg·m/s)÷(300 kg)

v = 2.25 m/s

• 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.

• When playing with a paddleball, the less time the ball is in contact with the paddle, the more force applied to the ball.

• If time is extended, less force will be applied to the object.

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.

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

by 3000 N of force acting for by 3000 N of force acting for 10.0 s (normal stop).10.0 s (normal stop).

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.

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

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)

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)

• 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

Motion PSc.1.1 OBJECTIVE: Understand motion in terms of speed, velocity, acceleration, and momentum.

selected reference point

total distance

reference pointin

reference pointfor

nonmoving point

reference pointmotionan

total displacement

displacement vectors

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