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Unit 2 - Motion Chapter 3 - Distance and Speed Unit 2 - Motion 1 / 76

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Page 1: Unit 2 - Motion - Chapter 3 - Distance and Speed...Chapter 3 - Distance and Speed Unit 2 - Motion 1 / 76 Precision and Accuracy Precision is a measure of how closely individual measurements

Unit 2 - MotionChapter 3 - Distance and Speed

Unit 2 - Motion 1 / 76

Page 2: Unit 2 - Motion - Chapter 3 - Distance and Speed...Chapter 3 - Distance and Speed Unit 2 - Motion 1 / 76 Precision and Accuracy Precision is a measure of how closely individual measurements

Precision and Accuracy

Precision is a measure of how closely individual measurements agree withone another.

Accuracy refers to how closely individual measurements agree with thecorrect or “true” value.

Unit 2 - Motion 2 / 76

Page 3: Unit 2 - Motion - Chapter 3 - Distance and Speed...Chapter 3 - Distance and Speed Unit 2 - Motion 1 / 76 Precision and Accuracy Precision is a measure of how closely individual measurements

Significant Figures

All digits of a measured quantity are called significant figures. Everymeasurement has uncertainty. Numbers in which there is no uncertaintyare called exact numbers and are rare in science.

What’s the temperature?

Unit 2 - Motion 3 / 76

Page 4: Unit 2 - Motion - Chapter 3 - Distance and Speed...Chapter 3 - Distance and Speed Unit 2 - Motion 1 / 76 Precision and Accuracy Precision is a measure of how closely individual measurements

Significant Figures

All digits of a measured quantity are called significant figures. Everymeasurement has uncertainty. Numbers in which there is no uncertaintyare called exact numbers and are rare in science.

What’s the temperature?

Unit 2 - Motion 3 / 76

Page 5: Unit 2 - Motion - Chapter 3 - Distance and Speed...Chapter 3 - Distance and Speed Unit 2 - Motion 1 / 76 Precision and Accuracy Precision is a measure of how closely individual measurements

Determining the Number of Significant Figures:

1 Zeroes between nonzero digits are always significant.

Ex: 1005 kg (four sig. figs.); 7.03 cm (three sig. figs.).

2 Zeroes at the beginning of a number are never significant.

Ex: 0.02 g (one sig. fig.); 0.0026 cm (two sig. figs.).

3 Zeroes at the end of a number are significant if the numbercontains a decimal point. If no decimal is present, the trailingzeroes are not significant.

Ex: 0.0200 g (three sig. figs.); 3.0 cm (two sig. figs.); 10 000 (onesig. fig.).

Unit 2 - Motion 4 / 76

Page 6: Unit 2 - Motion - Chapter 3 - Distance and Speed...Chapter 3 - Distance and Speed Unit 2 - Motion 1 / 76 Precision and Accuracy Precision is a measure of how closely individual measurements

Examples: Determine the number of significant figures in the followingmeasurements.

1 306 cm

2 2070 m

3 0.0065 m/s

4 0.350 km

5 3000 yr

6 4050 s

Unit 2 - Motion 5 / 76

Page 7: Unit 2 - Motion - Chapter 3 - Distance and Speed...Chapter 3 - Distance and Speed Unit 2 - Motion 1 / 76 Precision and Accuracy Precision is a measure of how closely individual measurements

Dimensional Analysis

Dimensional analysis is a systematic way of solving numerical problemsthat involve the conversion of units.

The strategy:

(((((

Given unit × desired unit

�����given unit

= desired unit

Unit 2 - Motion 6 / 76

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Examples: Use dimensional analysis to perform the following conversions.

1 35 centimetres to metres

2 2180 metres to kilometres

3 565 900 seconds to hours

Unit 2 - Motion 7 / 76

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Practice: Use dimensional analysis to perform the following conversions.

1 180 grams to kilograms

2 1.25 litres to millilitres

3 45 minutes to days

Unit 2 - Motion 8 / 76

Page 10: Unit 2 - Motion - Chapter 3 - Distance and Speed...Chapter 3 - Distance and Speed Unit 2 - Motion 1 / 76 Precision and Accuracy Precision is a measure of how closely individual measurements

Converting Units Using Two or More Conversion Factors

1 The average speed of a nitrogen molecule in air at 25◦C is 515 metresper second (m/s). Convert this speed to kilometres per hour.

2 Convert 20 feet per second to miles per hour. [1 mi = 5280 ft]

Unit 2 - Motion 9 / 76

Page 11: Unit 2 - Motion - Chapter 3 - Distance and Speed...Chapter 3 - Distance and Speed Unit 2 - Motion 1 / 76 Precision and Accuracy Precision is a measure of how closely individual measurements

What is distance?

Distance - The amount of space between two objects or points.

Measured in: kilometres (km), metres (m), centimetres (cm), ormillimetres (mm)

Unit 2 - Motion 10 / 76

Page 12: Unit 2 - Motion - Chapter 3 - Distance and Speed...Chapter 3 - Distance and Speed Unit 2 - Motion 1 / 76 Precision and Accuracy Precision is a measure of how closely individual measurements

What is time?

Time - The duration between two events.

Measured in: seconds (s), minutes (min), hours (h), or years (yr)

Unit 2 - Motion 11 / 76

Page 13: Unit 2 - Motion - Chapter 3 - Distance and Speed...Chapter 3 - Distance and Speed Unit 2 - Motion 1 / 76 Precision and Accuracy Precision is a measure of how closely individual measurements

Position - Is an object’s location relative to a reference point.

Uniform Motion - Motion at a constant speed in a straight line.

Inertia - The resistance of any physical object to any change in its state ofmotion.

Watch: “Motion”

Unit 2 - Motion 12 / 76

Page 14: Unit 2 - Motion - Chapter 3 - Distance and Speed...Chapter 3 - Distance and Speed Unit 2 - Motion 1 / 76 Precision and Accuracy Precision is a measure of how closely individual measurements

Investigating Speed, Time, and Distance

What is speed?

What is the relationship between speed, distance, and time?

→ Problem: Determine the average speed of a motorized toy car.

Unit 2 - Motion 13 / 76

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Relating Speed to Distance and Time

Average Speed

Speed - the rate (change in time) at which an object covers distance.

speed =distance

time

Average speed - the total distance divided by the total time for a trip.

vav =∆d

∆t

vav - “average speed”∆d = d2 − d1 (2nd distance - 1st distance)∆t = t2 − t1 (end time - start time)

Unit 2 - Motion 14 / 76

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Solve for the missing variable:

1 ∆d = 20.5 km; ∆t = 5.5 h

2 ∆t = 15.5 h; vav = 65.0 km/h

3 ∆d = 155 km; vav = 110 km/h

Unit 2 - Motion 15 / 76

Page 17: Unit 2 - Motion - Chapter 3 - Distance and Speed...Chapter 3 - Distance and Speed Unit 2 - Motion 1 / 76 Precision and Accuracy Precision is a measure of how closely individual measurements

Example 1. Jennifer walks to school a total distance of 4.5 km. The triptakes her 45.5 minutes. What was her average speed in km/h?

Unit 2 - Motion 16 / 76

Page 18: Unit 2 - Motion - Chapter 3 - Distance and Speed...Chapter 3 - Distance and Speed Unit 2 - Motion 1 / 76 Precision and Accuracy Precision is a measure of how closely individual measurements

Example 2. You are on a train and you see a sign that reads 120 km. Youdecide to measure the amount of time it takes to go from one sign toanother. If the signs are 10 km apart and it takes 391 s to travel betweenthem, how fast is the train going in km/h?

Unit 2 - Motion 17 / 76

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Example 3. Kara can predict how long it will take to bike from her houseto the beach. The distance is 45 km and she can bike at 20 km/h. Howlong will the trip take her?

Unit 2 - Motion 18 / 76

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Example 4. As a summer job, Mike analyzes grazing patterns of a herd ofbison. He notes that they graze at about 110 m/h for about 7.0h/d.What distance (in km) will the herd have travelled in two weeks (14 d).

Unit 2 - Motion 19 / 76

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Distance - Time Graphs

Graphs help us understand the relationship between two variables - an“independent” and a “dependent” variable.

Unit 2 - Motion 20 / 76

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In Distance - Time graphs:

Independent variable - time Dependent variable - distance

We can use these graphs to determine speed.

Which line below represents the fastest speed?

Unit 2 - Motion 21 / 76

Page 23: Unit 2 - Motion - Chapter 3 - Distance and Speed...Chapter 3 - Distance and Speed Unit 2 - Motion 1 / 76 Precision and Accuracy Precision is a measure of how closely individual measurements

Recall: y = mx + b is the general equation for a straight line.

y - dependent variable m - slope of the linex - independent variable b - y intercept of the line

Question: How can the slope of the best-fit line represent both vav =∆d

∆tand y = mx + b?

Unit 2 - Motion 22 / 76

Page 24: Unit 2 - Motion - Chapter 3 - Distance and Speed...Chapter 3 - Distance and Speed Unit 2 - Motion 1 / 76 Precision and Accuracy Precision is a measure of how closely individual measurements

Example: The motion of two bicycle riders, Tom and Jerry, is describedon a distance - time graph.

1 From just looking at the graph, which rider has the greater speed?

2 Calculate the speed of each rider by determining the slope of eachline.

0  

20  

40  

60  

80  

100  

120  

140  

160  

180  

200  

0   2   4   6   8   10   12   14   16  

Unit 2 - Motion 23 / 76

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Practice: Sketch a distance - time graph for a car that starts at rest andreaches a final speed of 80 km/h.

Unit 2 - Motion 24 / 76

Page 26: Unit 2 - Motion - Chapter 3 - Distance and Speed...Chapter 3 - Distance and Speed Unit 2 - Motion 1 / 76 Precision and Accuracy Precision is a measure of how closely individual measurements

Example: A car is travelling across the Confederation Bridge. Thedistances and times are listed below. They include a short stretch of roadbeyond the end of the 12.9 km bridge.

1 Plot a distance - time graph. Draw a best-fit straight line.

2 Using the graph, find the distance travelled after 5.0 min.

3 What is the required time to cross the bridge?

4 What is the speed of the car in kilometres per hour?

Time (min) Distance (km)0.0 0.0

2.0 2.4

4.0 4.8

6.0 7.2

8.0 9.6

10.0 12.0

12.0 14.4

Unit 2 - Motion 25 / 76

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Unit 2 - Motion 26 / 76

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Chapter 4 - Displacement and Velocity

What is acceleration?

What is the relationship between acceleration, speed, and time?

→ Problem:

Determine at what time the tennis ball is accelerating, decelerating,or neither.

Determine the formula for calculating acceleration.

Unit 2 - Motion 27 / 76

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When speed is not constant, it may change slowly or rapidly.

Acceleration (a) is the rate of change in speed. It is calculated using thefollowing formula:

aav =∆v

∆t

If the change in speed is the same in equal intervals of time, then this iscalled constant acceleration.

Average acceleration (aav ) is the average rate of change in speed of anobject.

Watch: “Car Accelerating” Watch: “X2” (off-ride)Watch: “X2” (on-ride) Watch: “KingDa Ka” (on-ride)

Unit 2 - Motion 28 / 76

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What are the units for acceleration?

Example:∆v = km/h∆t = h

=⇒ aav =km/h

h= km/h2

Examples:

1 ∆v = 105 km/h ∆ t = 0.0200 h aav = ?

2 ∆v = ? ∆t = 1.5 h aav = 25 km/h2

3 ∆v = 13 m/s ∆ t = ? aav = 8 m/s2

Unit 2 - Motion 29 / 76

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What are the units for acceleration?

Example:∆v = km/h∆t = h

=⇒ aav =km/h

h= km/h2

Examples:

1 ∆v = 105 km/h ∆ t = 0.0200 h aav = ?

2 ∆v = ? ∆t = 1.5 h aav = 25 km/h2

3 ∆v = 13 m/s ∆ t = ? aav = 8 m/s2

Unit 2 - Motion 29 / 76

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Example 1: A powerful car can accelerate from 0 to 100 km/h in 6.0 s.What is its average acceleration? [No conversions are required].

Unit 2 - Motion 30 / 76

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Example 2: Myriam Bedard (Olympic skier) accelerates at an average 2.5m/s2 for 1.5 s. What is her average speed (in m/s) at the end of 1.5 s?

Unit 2 - Motion 31 / 76

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Example 3: A skateboarder rolls down a hill and changes his speed fromrest to 1.9 m/s. If the average acceleration down the hill is 0.40 m/s2, forhow long (in s) was the skateboarder on the hill?

Unit 2 - Motion 32 / 76

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Refining the Acceleration Equation

Often, when acceleration is involved, the initial speed is known. The initialspeed is often a nonzero value; hence, speed can be more formerly definedas:

∆v = v2 − v1

Then, our acceleration formula is:

aav =v2 − v1

∆t

Unit 2 - Motion 33 / 76

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Example 4: Kerrin is moving at 1.8 m/s near the top of a hill. 4.2 s latershe is travelling at 8.3 m/s. What is her average acceleration?

Unit 2 - Motion 34 / 76

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Example 5: A bus with an initial speed of 12 m/s accelerates at 0.62m/s2 for 15 s. What is the final speed of the bus?

Unit 2 - Motion 35 / 76

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Example 6: A snowmobile reaches a final speed of 22.5 m/s afteraccelerating at 1.2 m/s2 for 17 s. What was the initial speed of thesnowmobile?

Unit 2 - Motion 36 / 76

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Example 7: In a race, a car travelling at 100 km/h comes to a stop in5.0 s. What is the average acceleration?

Unit 2 - Motion 37 / 76

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Speed - Time Graphs for Acceleration

Recall: Acceleration is basically a change in speed over time. Inspeed-time graphs, the following properties are important:

slope =rise

run=

speed

time

area under the line = distance travelled during that time interval

Unit 2 - Motion 38 / 76

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Practice: Find the area under the following curves.

Unit 2 - Motion 39 / 76

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Example 1: A boat on the St. Lawrence River travels at full throttle for1.5 h. From the area under the line of the speed-time graph, determinethe distance travelled. What was the average acceleration of the boat?

Unit 2 - Motion 40 / 76

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Example 2: Galileo rolls a ball down a long grooved inclined plane.According to a speed-time graph, what is the distance travelled in 6.0 s?What was the average acceleration?

Unit 2 - Motion 41 / 76

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Example 3: This is the speed-time graph of a train’s journey. What is thetotal distance travelled by the train?

Unit 2 - Motion 42 / 76

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Instantaneous Speed

Instantaneous Speed is the speed at a particular moment in time.

Example: The reading on a speedometer.

Note: For any object moving at a constant speed, the instantaneousspeed is the same at any time, and equals the constant speed.

Unit 2 - Motion 43 / 76

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Comparing Graphs:

Constant speed on Distance-Time graphs

Constant speed on Speed -Time graphs

Unit 2 - Motion 44 / 76

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Now suppose that speed is not constant; i.e. acceleration or decelerationoccurs. How do we find instantaneous speed?

We sketch the tangent to a point on a line. A tangent is a straight linethat just touches a curve at one point. This allows us to then calculate theslope of this line at this one point.

Unit 2 - Motion 45 / 76

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Example: Given the Distance (m) vs Time (min) graph below, answer thefollowing questions.

Unit 2 - Motion 46 / 76

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1) Which graph illustrates a constant speed for the whole trip?

2) Which graph shows a constantly changing speed?

3) Which graph(s) have an instantaneous speed of zero at some point?

4) What is the instantaneous speed at 3.5 min for each graph?

5) Calculate the average speed for 0 to 5.0 min for each cyclist.

Unit 2 - Motion 47 / 76

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Practice: What is the instantaneous speed at 2 seconds (time is on thebottom)? What is the instantaneous speed at the point (1, 3)?

Unit 2 - Motion 48 / 76

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Chapter 5 - Displacement and Velocity

Vectors: Position and Displacement

Reporting distances and speed without a direction is often not very useful.All distances and directions are generally stated relative to a referencepoint (origin or starting point).

Your position is the separation and direction from a reference point.

Figure: When we travel, Regina is our reference point.

Unit 2 - Motion 49 / 76

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We will describe distances and direction together using vectors. A vectorquantity is a quantity that involves a direction, such as position.

Examples: 2 km [E], 73 m [N], or 292 km [S].

Representation:

Unit 2 - Motion 50 / 76

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Displacement

Distance is a scalar quantity (involves only size). Displacement isdefined as the change in position.

Symbol: ∆~d

Displacement is usually calculated by the following:

∆~d = ~d2 − ~d1

Unit 2 - Motion 51 / 76

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Example: The Roughriders move on a straight downfield pass from theArgonauts’ 45-yard line to the Argonauts’ 20-yard line. Then (in classicfashion), lose 5 yards on the next running play.

Using the Argonauts’ 45-yard line as the reference point, what is the ball’s:

a) position after the pass? c) final displacement

b) final position? d) total distance travelled?

Unit 2 - Motion 52 / 76

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Adding Vectors Along a Straight Line

We can add straight-line vectors by either drawing a vector diagram or byusing arithmetic. We must draw our vectors to scale.

Note: ∆~dR - denotes the resultant vector

Example 1: Anne takes her dog for a walk. They walk 250 m [W] andthen back but only 215 m [E]. Find their resultant displacement.

Unit 2 - Motion 53 / 76

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Note: When computing vector addition algebraically, let [E] be positiveand [W] be negative. Then,

∆~dR = ∆~d1 + ∆~d2

Example 2: Anne goes for another walk. She leaves home and walks250 m [W] and then back back 175 m [E]. Find ∆~dR .

Unit 2 - Motion 54 / 76

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Example 3: Anne walks 30 m [W], stops to chat, then continues 50 m[W] before returning 60 m [E]. Find ∆~dR .

Unit 2 - Motion 55 / 76

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Adding Vectors at an Angle

Displacement can occur in all directions. We will use angles to specifymost displacements.

Method: If the direction that we’re interested in does not exactly matchone of the compass points, we write it as an angle from the closestcompass point.

Examples: a) [30◦ E of S] b) [10◦ E of N]

Unit 2 - Motion 56 / 76

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Example 1: Denise walks to Sarah’s home by going one block east andthen one block north. Each block is 160 m long. What is Denise’s finaldisplacement?

Unit 2 - Motion 57 / 76

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Example 2: This time, Denise walks 180 m east to get to Sarah’s home.Confused, she then walks 150 m west. What is Denise’s finaldisplacement?

Unit 2 - Motion 58 / 76

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Velocity

Speed is a very common quantity (car speed, airplane speed, etc.). Butoften speed is only useful when associated with some direction (windspeed direction, airplane flight patterns, etc.).

Velocity is a speed along with a direction.

~v =∆~d

∆tor ~vav =

∆~dR∆t

where,

∆~d or ∆~dR - change in displacement or resultant displacement

∆t - change in time

~v or ~vav - velocity or average velocity

Unit 2 - Motion 59 / 76

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Example 1: A train travels at a constant speed through the countrysideand has a displacement of 150 km [E] in a time of 1.7 h. What is thevelocity of the train?

Unit 2 - Motion 60 / 76

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Example 2: Monarch butterflies migrate from Eastern Canada to centralMexico, a resultant displacement of about 3500 km [SW] in a time ofabout 91 d. What is their average velocity in km/h?

Unit 2 - Motion 61 / 76

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Practice: A monarch butterfly usually flies during the day and rests atnight. If its velocity is 19 km/h [S] for 230 km [S] on one part of itsjourney to Mexico, how long does this take?

Unit 2 - Motion 62 / 76

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Example 3: A student travels 6.0 m [E] in 3.0 s and then 10.0 m [N] in4.0 s. Calculate the student’s average velocity.(Hint: Draw a vector diagram to determine the resultant displacement).

Unit 2 - Motion 63 / 76

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Practice 1: A car travels 8.0 km [N] and then turns suddenly west andtravels an additional 6.0 km. When the car stops, it has travelled a totaltime of 20.5 minutes. Calculate the car’s average velocity.

Unit 2 - Motion 64 / 76

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Practice 2: A cougar moves rapidly 2.0 km [S] in 5.0 minutes and thenmakes a sudden turn to the east 4.0 km, which takes 12.0 minutes.Determine the cougar’s average velocity.

Unit 2 - Motion 65 / 76

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Chapter 6 - Velocity and Acceleration

Position - Time Graphs

What’s the difference between a position - time graph and a distance -time graph?

Shows motion with constant speed Shows motion eastward withconstant velocity

Unit 2 - Motion 66 / 76

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Example 1: When Donovan Bailey finished the 4 x 100 m relay race, theteam’s time was 37.69 s, as describe graphically in the figure below.Determine Bailey’s velocity algebraically and graphically.

Unit 2 - Motion 67 / 76

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Note: The slope of the tangent at a point on a position - time graphyields the instantaneous velocity.

Example 2: A boat accelerates uniformly for seven seconds. What is theinstantaneous velocity at 4.9 s? Assume that east is positive.

Unit 2 - Motion 68 / 76

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Example 1: From the graph below, describe what is occurring in eachsegment.

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Velocity - Time Graphs

Recall: ~a =∆~v

∆t=⇒ on a velocity - time graph, slope = acceleration.

Example 2: What is the acceleration of the diving kingfisher (shown inthe graph below)? Note: Up is the positive direction and down is negative.

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Acceleration and Velocity

Recall from Chapter 2: a =v2 − v1

∆t

Vector acceleration is simply change in velocity in a given time:

~a =~v2 − ~v1

∆t

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Example 1: Suppose a plane (taking off) starts from rest and acceleratesto a final velocity of 270 km/h [E] in a time of 32 s. Calculate theacceleration of the airplane. Assume east is positive.

Watch: “Very fast take off”

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Example 2: Suppose the same plane reaches its destination and touchesdown on the runway travelling at 305 km/h [E]. If the plane takes 25 s tocome to a complete stop, what is its vector acceleration?

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Example 3: An air puck on an air table is attached to a spring. The puckis fired across the table at an initial velocity of 0.45 m/s [right] and thespring accelerates the air puck at an average acceleration of 1.0 m/s2

[left]. What is the velocity of the air puck after 0.60 s? Assume that rightis positive and left is negative.

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Investigating Acceleration due to Gravity

What is the acceleration of falling objects?

→ Problem: The acceleration of gravity was first discovered by IsaacNewton. We will attempt to make him proud by discovering what thisacceleration is. We will use his formula to help us:

ag =2h

t2where,

ag - acceleration due to gravity

h - height

t - total time

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Example 1: A person throws a ball straight up from the ground. The ballleaves the person’s hand with an initial velocity of 10.0 m/s [up]. Assumeup is positive and down is negative.

a) What is the velocity of the ball after 0.50 s?b) What is the velocity of the ball after 1.50 s?

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