Science 1206 Physics Exploits Valley High

Science 1206 Physics Exploits Valley High

Introduction to Motion

Motion is the act or process of changing position or place.

Examples of moving objects.

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Science 1206 Physics Exploits Valley High

ski-doo

car -

train

softball

Olympics – methods of determining time

Distance (d)

The amount of space between two objects or points

Units: miles, kilometers, meters, feet

The most common unit of distance is the meter

Time (t)

Duration between two events

Usually measured in seconds, minutes or hours

Speed (v)

The rate of change of distance.

Example: the ski-doo is traveling 40 km/hr

Advantages and disadvantages of increasing speed on highways

Advantages

less time in traveling

shorted journey times may mean less driver fatigue

vehicles are on the highways for shorter periods of time, reducing congestion

high highway speeds, so long as all vehicles are traveling equally fast and road conditions are good, result in no more accidents than lower speeds

Disadvantages

accidents at high speed tend to be more serious

high-speed vehicles need more space between them

traveling fast uses considerably more fuel

Conditions are frequently less than ideal on Canadian roads, so slower driving would result in fewer accidents

lower speed limits might even persuade some people to take less polluting forms of public transportation

Technologies designed to prevent excessive speeds on highways

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Science 1206 Physics Exploits Valley High

photo-radar

highway patrol cars

speed traps

Measurement and Calculations

A measurement is the dimension, quantity, or capacity determined by measuring.

Three criteria for successful measurements

Measuring device

Measuring methodology

Accurate reading

Measurement device(s)

Mass – scale

Time – watch

Temperature – thermometer

Voltage – volt meter

Electric Current – volt meter

Significant digits.

In science, it is important that the degree of uncertainty in measurements be shown. One method of indicating uncertainty is by the number of significant digits recorded.

Significant digits are those digits obtained from a properly taken measurement. Significant digits as obtained from a measurement are all of the certain digits from a measurement plus one uncertain digit.

Rules for counting significant digits:

1. Count all digits from 1 to 9 plus zeros in between and following other digits;
2. Do not count zeros in front of a value because they only serve to set the decimal place.

3. Zeroes at the end of a number and to the right of the decimal point are significant.

Example:

Measured Value / # sig. digits / Measured Value / # sig. digits
156 g / 3 / 120.50 L / 5
0.2602 m / 4 / 0.05002 s / 4
6.02  1023 molecules / 3 / 7.2 oC / 2
134.56 kg / 5 / 45.03 ml / 4

Rules for addition and subtraction:

When adding and subtracting, add or subtract, then round off the answer to the least number of decimal places contained in the question. The answer cannot be any more precise than the least precise part of the question

Rules for multiplication and division:

1. Multiply or divide and then round-off to the least number of significant digits found in the question.
2. When multiplying or divide an uncertain value by an exact number, the answer has the same precision as the value with uncertainty.
3. When multiplying or divide a value with uncertainty in order to use a different SI prefix, the original significant digits are retained identically.

Practice:

Measured Value

/ # s.d. / Calculated Value / Rounded-off (# s.d.)
12.42 / 0.1495 / (3)
0.1407 / 29.95 / (3)
10.0 / 139.49 / (3)
1000 / 10.54 / (3)
0.060 / 100.4 / (3)
126 / 9.998 / (2)
15.00 / 80.46 / (2)
0.0004 / 197.042 / (4)
40 / 0.0462 / (2)
0.100001 / 8.29 / (2)

Accuracy and precision in measurement

Precision – The place value of the last digit obtained from a measurement or calculation; indicated by the number of decimal places.

Ex.5.1Least precise

5.13

5.135 Most precise

Accuracy – A comparison of an experimental value with an accepted value, usually expressed as a percent error.

Archer shooting at a target:

precise, but not accurate

good accuracy but poor precision.

Archer is both accurate and precise.

Results are both accurate and precise with the exception of an obvious error. Because several measurements were made, we can ignore the error as an obvious mistake, probably due to the archer’s error.

Sources of Error

No measurement is exact; any scientific investigation will involve error.

1. Random error (uncertainty)

• an error that relates to reading a measuring device
• eg. a person measuring the length of an object using a ruler must estimate the last digit; another person may not estimate to the same digit
• can be reduced by taking many measurements and then averaging them (and having the same person take the measurement each time)

1. Systematic error

• An error due to the use of an incorrectly calibrated measuring device.
• Eg. a clock that runs slow or a ruler with a rounded end
• Can be reduced by inspecting and recalibrating equipment regularly

1. Parallax

• The change in relative position of an object with a change in the viewing angle
• Eg. the view of a speedometer is not the same to the driver as it is to a passenger
• Can be reduced by taking readings from an eye-level, head on perspective

1. Discrepancy

• The difference between the value determined by your experimental procedure and the generally accepted value

% Discrepancy = Experimental value – accepted value X 100

accepted value

Example: A student measures the acceleration due to gravity and finds it to be 9.72 m/s2. If the accepted value is 9.81 m/s2, what is the percent discrepancy.

% Discrepancy =

EXERCISES:Calculate the percent discrepancy for each of the following:

1. A group of students estimate the speed of light at 2.4 x 108 m/s. The accepted value however is 3.00 x 108 m/s.

2.An science student measures the acceleration due to gravity to be 9.75 m/s/s. The expected value however is 9.8 m/s/s.

Changing numbers from scientific notation to standard notation.

Example 1:

Change 6.03 x 107 to standard notation.

remember, 107 = 10 x 10 x 10 x 10 x 10 x 10 x 10 = 10 000 000

so, 6.03 x 107 = 6.03 x 10 000 000 = 60 300 000

answer = 60 300 000

Instead of finding the value of the base, we can simply move the decimal seven places to the right because the exponent is 7.

So,6.03 x 107 = 60 300 000

Example 2

Change 5.3 x 10-4 to standard notation.

The exponent tells us to move the decimal four places to the left.

so, 5.3 x 10-4 = 0.00053

Example 3

Change 56 760 000 000 to scientific notation

Remember, the decimal is at the end of the final zero.

The decimal must be moved behind the five to ensure that the coefficient is less than 10, but greater than or equal to one.

The coefficient will then read 5.676

The decimal will move 10 places to the left, making the exponent equal to 10.

Answer equals 5.676 x 1010

Example 4

Change 0.000000902 to standard notation

The decimal must be moved behind the 9 to ensure a proper coefficient.

The coefficient will be 9.02

The decimal moves seven spaces to the right, making the exponent -7

Answer equals 9.02 x 10-7

Uniform motion

Uniform motion – Movement in a straight line at a constant speed

Non-uniform motion ( accelerated motion) – Movement involving changes in speed or direction or both.

Investigating Speed

Speed is the rate of change of motion.

Average speed is the total distance divided by the total time for a trip

Instantaneous speed – The speed of an object at a particular instant, it is not affected by its previous speed or by how long it has been moving

Change km/h to m/s using shortcut rule

Shortcut: 1 m/s = 3.6 km/h

Example 1: Change 4 m/s to km/h

Example 2: Change 5.6 km/h to m/s

Example 1:

Eiko skeates to school, a total distance of 4.5 km. She has to slow down twice to cross busy streets, but overall the journey takes her 0.62 h. What is Eiko’s average speed during the trip?

Example 2:

A car is moving at a constant speed of 88 km/h when a dog suddenly appears on the road ahead. The driver immediately brakes to avoid hitting the dog.

(a)Convert 88 km/h into meters per second

(b)If the reaction time of the driver is 0.2 s, how far has the car moved by the time the driver just touched the brake pedal?

Example 3:

Johnny walks to school, a total distance of 6.4 km. If the trip takes him 48 minutes, what is his average speed in km/h?

Example 4:

In 1979, Bryan Allen pedalled the Gossamer Albatross aircraft 35.0 km across the English Channel in a time of 169 minutes.

(a) Calculate the average speed of the aircraft in km/h.

(b) Assuming that he maintained this same average speed, what total distance could he cover in 5.3 hours?

Answer

(a) ∆d = 35.0 km

∆t = 169 min

vav = ? km/h

∆t = 169 min x 1 h = 2.82 h

60.0 min

vav = ∆d

∆t

= 35.0 km

2.82 h

= 12.4 km/h (rounded to 3 significant digits)

The average speed of the aircraft was 12.4 km/h.

(b) ∆d = vav ∆t

= 12.4 km/h x 5.3 h

= 66 km (rounded to 2 significant digits)

Assuming he maintained the same average speed, then he would cover a total of 66 km.

Example 5:

Susan and Margie are competing in a 10.0 x 103 m race. Susan can run at 2.0 m/s while Margie can run at 1.5 m/s.
a)How long will it take each person to finish the race?
b)After Susan crosses the finish line, how long will it be before Margie crosses?

a) For Susan,
t = d/v
= 10.0 x 103 m/2.5 m/s
= 4000 s
= 4.0 x 103 s
= 66.7 min
For Margie,
t = d/v
= 5.0 x 103m / 1.6 m/s
= 3125 s
= 3.13 x 103 s
= 52.1 min

b) Since Susan crosses in 4000 s,
she is 4000 s – 3125 s = 875 s or 14.6 min
faster than Margie.

Example 6:

A bullet is shot from a rifle with a speed of 720.0 m/s.What time is required for the bullet to strike a target 3240.0 m to the east?

Given the formula v = d/t and any two of speed, distance or time, calculate the unknown value

d / t / v
4.0 km / 3.1 s
4.0 km / 6.2 km/h
3.0 s / 3.5 m/s

Graphing Uniform Motion

Advantage of using graphs to communicate information

Graphs help us understand the relationship between two variables.

We can see whether the dependent variable increases or decreases with the independent variable.

On a distance-time graph

• Time is the independent variable
• Distance is the dependent variable

Investigating distance vs time graphs

Draw the d vs t graph given a data table. Plot the independent variable on the horizontal (x) axis

Example 1: Consider a running white-trailed deer

Time (s) / Distance(m)
0 / 0
1 / 10
2 / 20
3 / 30
4 / 40
5 / 50
6 / 60

Determine the slope of a graph line.

At what point does the graph cross the vertical axis (d) ? ______

Explain how the formula y = mx + b is used to describe a graph

Consider: d = mt + b

Example 2:

Consider; A hummer traveling on a country road

Time (s) / Distance(m)
0 / 0
1 / 1
2 / 1.8
3 / 2.9
4 / 4.2
5 / 5.0
6 / 5.9
7 / 6.9
8 / 8.2
9 / 9.2
10 / 10.0

Plot the graph and draw the line of best fit.

Speed is determined from the slope of the best-fit straight line of a distance-time graph

In summary, the speed of an object in motion can be determined from the slope of a distance-time graph.

Example 3:

Use the d vs t graph of different objects to determine the fastest or slowest motion.

Time (s) / FreeStyle (m) / ButterFly (m)
0 / 0 / 0
5.0 / 9.5 / 8.2
10.0 / 18.5 / 16.2
15.0 / 27.0 / 24.7
20.0 / 35.5 / 32.8
25.0 / 45.0 / 42.0

Construct both graphs on the same axis and draw the line of best fit.

From the regression line determine the speed for each type of swimming.

Perform a lab experiment (lab #7) to determine the average speed of an object

Investigating speed vs time graphs

Draw the v vs t graph for the motion of an object

Time (s) / Velocity (m/s)
1 / 10
2 / 10
3 / 10
4 / 10
5 / 10
6 / 10
7 / 10

What do you notice about the graph? and what does it mean?

Determine the area under the line of the v vs t graph for uniform motion using area = L x W.

Explain the importance of the area calculation

In summary, the area under a speed vs time graph for uniform motion is equal to the distance.

Vectors

Often when dealing with motion, it is necessary to give your position relative to a reference point.

Example:

Rescue

Using GPS to find your direction of travel.

All positions are generally stated relative to a reference point, usually the starting point.

Position is defined as the separation and direction from a reference point.

Example – you may be at a position 152 m [W]  one hundred and fifty two metres west of the reference point

The direction is given in square brackets and is usually a compass direction or a right/left or forward/backward direction

On a straight line, position is sometimes stated as a positive or negative value relative to a zero point

Distinguish between scalar and vector quantities

Scalar Quantity – A quantity that involves only size, but not direction

Vector Quantity – A quantity that involves both size and direction

Quantity Symbol / Example
Scalar Quantity
Distance / ∆d / 292 km
Time / ∆t / 3.0h
Speed / v / 100 km/h
Vector Quantity
Displacement / / 292 km [S]
Velocity / / 50 km/h [N]

Vector

A vector is a line segment that represents the size and direction of a vector quantity.

Vectors can be drawn as a sketch or drawn to scale.

Example: 15 km [E]

Example : 19 m [S]

Follow these rules when drawing a single vector:

State the direction – usually using N, S, E or W as reference

Draw a line to the stated scale or write the size of the vector next to the line

The direction of the line represents the direction of the vector and the length of the line represents the size of the vector

Multiple vectors

Use the head-to-tail rule

• Join each vector by connecting the ‘head’ end of one vector to the ‘tail’ end of the next vector

This sum is called the resultant

• Find the resultant by drawing an arrow from the tail of the first vector to the head of the last vector

Displacement a vector quantity is thechange in an objects position ().

This includes the ∆ (delta) for change and d for position

∆d = d2 – d1  the final position minus the initial position

Displacement is a vector quantity and is communicated as the size of the quantity and the direction.

It represents the straight line distance from some initial position in a given direction to a final position

Calculating the displacement(one dimensional).

We must assign positive or negative directions to the value of the quantity

Chose one direction as positive and the other will automatically be negative

Examples :

1.Walk 15 km [North] and then 35 km [South].

Total Distance ______Displacement ______[ ]

1. An object moves 15 km [E] , then 8 km[W] and finally 12 km [E] before stoping.

Total Distance ______Displacement ______[ ]

3.A car travels 8 km [N] then 14 km [S] and finally 20 km [N] .

Total Distance ______Displacement ______[ ]

4.A student walks 5.0 km [S] and then 3.0 km [N].

5.0 km [S] + 3.0 km [N] = 5.0 km [S] + - 3.0 km [S]

= 2.0 km [S]

The student walks a distance of 8.0 km and the student’s displacement is 2.0 km [S].

Displacement two dimensions.

When vectors act at an angle to each other, a scale drawing can be used to determine resulting vector.

Note the steps to follow (on page 427) when drawing vector diagrams. Mathematics can also be applied when determining the resulting vector in vector diagrams. If vectors form a right triangle then Pythagoras theorem can be applied to find the magnitude of the resulting vector.

Trigonometry can be used to help us determine the direction of the resulting vector.

Example 1:

Eric leaves the base camp and hikes 11 km, north and then hikes 11 km east. Determine Eric's resulting displacement.

The result of adding 11 km, north plus 11 km, east is a vector with a magnitude of 15.6 km.

Example 2:

/ R2 = (5)2 + (10)2
R2 = 125
R = 11.2 km

Example 3:

/ R2 = (30)2 + (40)2
R2 = 2500
R = 50 km

Velocity

Velocity is the rate of change of displacement. It is a vector quantity.

Example 1

A train travels at a constant speed through the countryside and has a displacement 150 km [E] in a time of 1.7h. What is the velocity of the train?

∆d = 150 km [E]

∆t = 1.7 h

v = ?

v = ∆d/∆t

= 150 km[E]/1.7h

= 88 km/h [E]

Example 2

A jogger travels 52 m[E] in 10.0 s and then 41 m [W] in 8.0 s for a total distance of 93 m in a total time of 18.0 s.

Average speedAverage Velocity

Vav = ∆d/∆tVav = ∆dR/∆t

= 93 m /18.0s = 11 m [E]/18.0 s

= 5.2 m/s = 0.61 m/s [E]

Example 3:

Monarch butterflies migrate from Eastern Canada to central Mexico, a resultant displacement of 3500 km [SW] in a time of 91 days. What is the average velocity of the butterflies in kilometers per hour?

∆dR = 3500 km [SW}

∆t = 91 d = 91 d x 24h/d = 2184 h

Vav= ∆dR/∆t

= 3500 km [SW]/2184h

= 1.6 km/h [SW]

Example 2:

A monarch butterfly usually flies during the day and rest at night on its migration. If a particular butterfly is traveling at an average velocity of 19 km/h [S] for 230 km [S] on one part of its journey to Mexico, how long does this take?

∆d = 230 km [S]

Vav = 19 km/h [S]

∆t = ?

Vav= ∆dR/∆t

∆t = ∆dR/∆ Vav

= 230 km[S]/19 km/h [S]

= 12 h

Example 3:

Determine the velocity of the object for each of the intervals.

Example 4:

Describe the motion of the jogger on each section of the velocity-time graph. Be sure to specify the direction of the motion.

Answer

A- The jogger runs north at a constant velocity of 3.0 km/h for about 1.0 h.

B- The jogger slows down from 3.0 km/h to 0 km/h in 1.0 h.

C- The jogger remains stationary for another hour, no velocity.

D- The jogger speeds up in the south direction from 0 km/h to –4 km/h for 1.0 h.

E- The jogger remains stationary for another hour, no velocity.

F- The jogger speeds up in the north direction from -4 km/h to 0 km/h for 1.0 h.

Acceleration

Acceleration (a) is the rate of change in speed ,

It is calculated by the ration of the change in speed ∆V to the time interval ∆t

Units m/s2

Constant Acceleration

The same change in speed occurs in each equal interval of time.

Example: Free fall. g = 9.8 m/s2

t (s) / V (m/s)
0 / 0
1 / 9.8
2 / 19.6
3 / 29.4

Example 1

Suppose you speed up on a motorcycle from rest (0 m/s) to 9.0 m/s on a time of 2.0 s. What is the objects acceleration.

Example 2:

What is meant by an acceleration of 4.5 m/s2

You will increase your speed by an average of 4.5 m/s during every second of travel

Starting from rest, your speed is 4.5 m/s at the end of the 1st second

9.0 m/s at the end of the 2nd second

13.5 m/s at the end of the 3rd second

Example 3:

An F-15 jet fighter starts from rest and reaches a speed of 330 m/s in 2 seconds.What is the planes acceleration?

Example 4:

A physics student drops a rock into the well. 4.5 seconds after the rock is dropped the student sees it hit the bottom. The rock accelerates downwards at 9.80 m/s2. How fast is the rock traveling the instant before it hits the bottom?

Example 5:

A bicyclist travels from 15.6 m/s to 21.1 m/s in 3.0 s. What is the acceleration of the bicyclist?

Example 6:

A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the acceleration of the car.

Example 7:

A race car decelerates uniformly from 46.1 m/s to 18.5 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled.

Graphing Accelerated motion

Example 1:

Draw a speed vs time graph for the data table below.

Acceleration on a bicycle

Time (s) / Speed (m/s)
0 / 0
2 / 5
4 / 10
6 / 15
8 / 20
10 / 25
12 / 30

Determine the slope of a v-vs-t graph.

In summary, the slope of the line on a speed-time graph provides the acceleration.

Example 2: Construct a speed time graph for the acceleration of a bicycle below.

Time(s) / Speed (m/s)
0 / 0.0
10.0 / 2.0
20.0 / 4.0
30.0 / 6.0
40.0 / 8.0
50.0 / 10.0

Calculate the area under the graph.

The area under the line in a speed-time graph equals the distance traveled during the time interval.

Example 3:

Draw an acceleration time graph for the following speed time graph.

Acceleration for t = 0 – 15 s
Acceleration for t = 15 – 25 s
Acceleration for t = 25 – 40 s

Distance-time graph for accelerated motion

Example 1:

Construct a distance time graph for the motion of a parachutist who is freefalling and accelerating at 9.8 m/s2

t (s) / d ( m)
0 / 0.0
1 / 4.9
2 / 19.6
3 / 44.1
4 / 78.4
5 / 122.5

What do you notice about the graph?

The distance time graph for a uniformly accelerated motion is a curved line.

Example 2:

Construct a distance time graph for the first 5.0 s of a car traveling at 100 km/h ( 28 m/s) that starts decelerating at 5.0 m/s2 .

t (s) / d ( m)
0 / 0.0
1 / 25
2 / 46
3 / 61.5
4 / 72
5 / 77.5