Science 10 Physics Unit B Read pg. 465 - 477. Conversions Review.

Science 10 Physics

Unit BRead pg. 465 - 477

Conversions Review

How to Multiply Fractions

• 3 x 4 = 1 5

• 3 x 7 = 8

• 3 x 5 = 9

12 5

21 8

15 9

Unit Conversions (Distance)

• E.g. 3.75km m 1 km = 1000 m

• Try: 5.85m m 1 km = 1000 m

3.75 km x 1000 m = 1 km

5.85 km x 1000 m = 1 km

3750 m

5850 m

• E.g. 427cm m 100 cm = 1 m

• Try: 865 cm m 100 cm = 1 m

427cm x 1 m = 100cm

865 cm x 1m = 100cm

4.27 m

8.64 cm

• E.g. 67 mm m 1000 mm = 1 m

• Try: 765 mm m 1000 mm = 1 m

67 mm x 1 m = 1000 mm

765 mm x 1m = 1000 mm

0.067 m

0.765 m

• E.g. 580 m km 1000 m = 1 km

580m x 1 km = 1000 m

0.580 km

Try Unit Conversions (Time)• E.g. 2.75 h min 1 hr = 60 min 2.75 h x 60 min = 165 min 1 h

• Try: 42 min h 42 min x 1h = 0.70 h 60 min

• E.g. 2.10 h s1 hr = 3600 s2.10 h x 3600 s = 7560 s

1h

• Try: 3 h s 3 h x 3600s = 10800 s 1 h

• Units to know for this unit:• Distance, height = meters (m)• Time = seconds (s)• Speed, velocity = meters per

second (m/s)• Acceleration = meters per second2

(m/s2)• Work, energy = Joules (J)• Force = Newtons (N)• Mass = kilograms (kg)• Efficiency = percent (%)

• Formulas to know for this unit: v = d t vave = vi + vf

2 a = vf - vi

t F = ma W = Fd Ep = mgh Ek = 1/2 mv2

% Efficiency = useful output x 100% total input

Rearranging formulas

• You need to ISOLATE the variable you are trying to solve for

• What ever mathematical operation you do to one side of the = you need to also do to the other side

• Ex. v = d Solve for d and solve for t t

Drawing Graphs

Parts of a Graph

• All graphs should have:– A horizontal axis (or x axis, ALWAYS TIME!)– A vertical axis (or y axis)– A title– Labels on each axis– Units for each axis– Appropriate scale (numbering on both axis)

Example:

Distance-Time Graph

Label (Units)

Label (Units)

Title

Scale

*Note: In Physics, time will always be the horizontal axis

vave = vi + vf 1. solve for vi

2 2. solve for vf

a = vf - vi 3. solve for vi

t 4. solve for t F = ma 5. solve for a W = Fd 6. sove for F Ep = mgh 7. solve for h Ek = 1/2 mv2 8. solve for v % Eff = useful output x 100% 9. solve for total input

total input 10. solve for total output

Rearrange the following formulas:

Draw the following graphTime (s) Distance (m)

0.0 0.02.0 5.04.0 10.06.0 15.08.0 20.0

10.0 25.012.0 30.014.0 35.016.0 40.0

1.1 MotionRead pg.126-131

Energy

• Causes changes in the motion to occur to an object

• It can speed objects up, slow them down or change their direction

Uniform Motion

• Describes a type of movement

• It occurs when an object travels in a straight line at a constant speed

• is difficult to maintain so ….we use AVERAGE SPEED

v = d t

= changed = distance in m or kmt = time in s or hv = speed in m/s or km/h

Average Speed = distance traveled time

v = d t

v = dfinal – dinitial

tfinal – tinitial

v = speed (m/s or km/h)d = change in distance (m or km)t = change in time (s or h)

Example 1 A baseball travels 200 m in 1.50

seconds. What is the average speed of the baseball?

d = 200 m v = ?t = 1.50 s

v = 200m 1.50 s

v =

v = 133 m/s

________ ________

Δd

Δt

a.) A baseball travels 20.0 m in 1.50 seconds. Calculate the average speed.

v = 20.0 m 1.50 s v = 13.3 m/s

d = 20.0 m

t = 1.50 s

v = ?

Try:

Δd

Δtv =

b.) If Lance Armstrong bikes 200.0 m in 10.0 s, what is the cyclist’s average speed?

v =Δd

Δt

200.0 m

10.0 s

20.0 m/s

d = 200.0 m

t = 10.0 s

v = ? v =

v =

c.) If a train traveled 100 km in 0.500 hours what is its speed in km/h and in m/s?

d = 100 km

t = 0.500 h v = ?

100 km 0.500 h 200 km/h

Δd

Δtv =

v =

v =

200 km x 1000 m/km

1 h x 60 min/h x 60 sec/min

200 km/h 3.6

= 55.6 m/s =

Example 2

A car travels 1.00 km at a constant speed of 15 m/s. What time is required to cover this distance?

d = 1.00 km = 1000 m

v = 15 m/st = ?

t = d v t = 1000m 15 m/s

v =

t = 67 s

________

________

Δd

Δt

Try:a.) How long would it take a car to travel

4000 m if its speed was 40.0 m/s?

d = 4000 m

v = 40.0 m/s

t = ? t = d v = 4000 m 40.0 m/s

= 100 s

v =Δd

Δt

b.) How long would it take a car to travel 2000 m if its speed was 10.0 m/s?

d = 2000 m

v = 10.0 m/s

t = ? t = d v = 2000 m 10.0 m/s

= 200 s

v =Δd

Δt

c.) How long would it take a car to travel 8000 m if its speed was 30.0 m/s?

d = 8000 m

v = 30.0 m/s

t = ? t = d v = 8000 m 30.0 m/s

= 267 s

v =Δd

Δt

Example 3

A motorist travels 406 km in 4 hours and 15 minutes. What is the average speed in km/h and m/s?

d = 406 km v= ?t = 4 hour + 15min 60 min/h = 4.25 h

v =

v = 406 km 4.25 hv = 95.5 km/h

= /3.6= 26.5 m/s

________ ________

Δd

Δt

v

Example 4. How far of a distance will a car cover if it travels 2.00 m/s for 1.00 min?

d = ?

v = 2.00 m/s

t = 1.00 min = 60.0 s

d = v t =2.00 x 60.0

= 120 m

v =Δd

Δt

• distance varies directly with time when speed is constant

• Have the following components:– time is the (x-axis)– distance is the (y-axis)– the slope of the line is the speed of an

object

Distance Time Graphs

• speed describes the rate of motion an object has

t(s) d(m)0 01.0 202.0 403.0 604.0 805.0 100

dist

ance

(m

) 2

0

40

60

80

100

0 1 2 3 4 5

time (s)

d

t

Distance-Time Graph

• The steepness of the graph is the slope

• Example:

slope = y2 – y1

x2 – x1

= 80m – 20m

4.0 s- 1.0 s

= 20 m/s

slope = rise run

= y2 – y1

x2 – x1

• The steeper the slope the higher the speed.

Dis

tanc

e (m

)

time (s)

Which line shows the greatest speed? The slowest speed?

A

B

C

Try the Following:

• Make a Distance time graph for the following

• Calculate the speed of the boat

v = =Δd

Δt30 m – 10 m

6.0 s – 2.0 s

= 5.0 m/sV

Speed- Time Graphs

• The area under the graph is the distance an object travels

• The slope of the line gives you information about the speed

• E.g. A slope of zero (flat line) = uniform motionUpward slope = speed is increasingDownward slope = speed is decreasing

Timet (s)

Speedv (m/s)

0.0 5.002.0 5.004.0 5.006.0 5.008.0 5.0010.0 5.00

0 2 4 6 8 10

Time (s)

Spe

ed (

m/s

)0

1

2

3

4

5

6

Uniform Motion

• Can be used to determine the distance an object travels…… calculate the area under the line

Example 1Calculate the area under the following speed-time graph up to 10.0 s.

Time (s)

Spe

ed

(m/s

)

10.0

5.0

0.0 5.0

10.0

SolutionCalculate the area under the following speed-time graph up to 10.0 s.

area = w A = (10.0 s)(5.0 m/s) A = 50.0 m

Time (s)

Spe

ed

(m/s

)

10.0

5.0

10.0

0.0

Example 2Calculate distance travelled by an object in 20.0s.

Time (s)

Spe

ed (

m/s

)

20.0

10.0

0.0

SolutionCalculate distance travelled by an object in 20.0s.

Time (s)

Spe

ed (

m/s

)

20.0

10.0

0.0

A= 1 b x h 2 A= 1 20 x 10 2

A = 100 m

Calculate distance travelled by an object in 40.0s.

Time (s)

Spe

ed (

m/s

)

20.0

10.0

0.0

Try the Following

40.0

20.0

30.0

Calculate distance travelled by an object in 40.0s.

Time (s)

Spe

ed (

m/s

)

20.0

10.0

0.0

Solution

40.0

20.0

30.0

A= 1 b x h 2 A= 1 40 x 25 2

A = 500 m

1.2 Velocity

Vectors Verses Scalar

• Scalar quantities:– involve only magnitude (amount)

• Vector quantities:– involve both magnitude and direction– Are drawn using arrows

E.g.) Speed = 20 m/s

E.g.) Displacement = 20 m [N]

Velocity = 20 m/s [N]

• when describing a vector, we have two quantities that indicate the direction:

from the

using

-Degrees x or y axis

-Degrees compass directions(N, S, E, or W)

E.g. 6 m [30°]

up (90°)

down (270°)

right (0°)

left (180°)

30°

6 m

E.g. 10 m [right]

10 m

8 m

40°

2 m

Cartesian Method

It uses N-S-E-W .. the angle is relative to E (east) or W (west), or a direction at N, S, E or W

N

S

EW

angle (°)

N of E

angle (°)

N of W

angle (°)

S of E

angle (°)

S of W

(-)

(-)

(+)

(+)(+)

(+)

(-)(-)

Navigator Method

N

E

S

W

+ y axis

y axis

+ x axis x axisA

B

10 km1.0 m/s 80

65

vector A:

vector B:

10 km, 80 E of the y axis10 km, 80 E of S

1.0 m/s, 26 S of the x axis1.0 m/s, 26 S of W

N

E

S

W30º

15º

55º 30° S of E

15°N of W

55°W of SS

y axis

+ y axis

+ x axis x axis

AB

C

Distance Verses Displacement

• Distance (d)– is how far an object travels– It is a scalar quantity (magnitude)

• Displacement (d)– Is change in both distance and direction – It is a vector quantity (magnitude and

direction)

Vectors Direction• Vectors in the Same Direction:

– To find the distance: Add them together

– To find the displacement: Add them together and include the

direction

10 m 5 m = 15 m

Example1

A person runs 25 m south and then another 15 m south.

(a) What is the distance travelled?  

    (b) What is the displacement? 

Solution

A person runs 25 m south and then another 15 m south.

(a) What is the distance travelled?  

    (b) What is the displacement? 

25 m

15 m40 m

40 m South

• Vectors in Opposite Directions:

– To find the distance: Add them together

– To find the displacement: take the difference between the

two numbers and include

the direction20 m

5 m

Distance = 25 m

Displacement = 15 m E

  Example 2

A plane flies 200 km north and then turns around and comes back 150 km.

a) What is the distance travelled? 

b) What is the displacement?

  Solution

A plane flies 200 km north and then turns around and comes back 150 km.

a) What is the distance travelled? 

b) What is the displacement?

350 km

50 km N

200km 150 km

50 km

Speed verses Velocity

• Speed (v):• is the rate of an objects motion• It is a scalar quantity • The formula is:

v= d t

v = speed d = distance (dfinal – dinitial) t = time

• Velocity ( )• Describes the rate of motion and the

direction of the object’s motion• It is a vector quantity• the formula is:

vave = d

t

vave = dfinal – dinitial

tfinal - tinitial

v

Example 3A student throws a boomerang north. It travels 35.0 m before it turns around and travels 33.0 m back to him. If the total flight of the boomerang took 5.00s, determine the following:

a) distance

b) displacement

c) speed

d) velocity

SolutionA student throws a boomerang north. It travels 35.0 m before it turns around and travels 33.0 m back to him. If the total flight of the boomerang took 5.00s, determine the following:

a) distance d = 35.0 m + 33.0 m = 68.0 m

b) displacement

2.0 m [N] c) speed

  v = d/t = 68.0 m/5.00 s = 13.6 m/s

d = 35.0 m [N] - 33.0 m [S] =

d) velocity   v = d/t = 2.0 m [N]/5.00 s = 0.40 m/s [N]

Example 4 A plane flies south to Edmonton International Airport, which is 465 km from the Fort McMurray Airport. If the flight takes 50.0 minutes what is the average velocity of the plane in km/h and m/s?

t = 50.0 min 60min/h = 0.8333…. h

558 km/h x 1000 = 155 m/s [S] 3600

vave = d t= 465 km [S] – 0 km 0.8333… h= 558 km/h [S]

Example 5A train travels at 12.0 m/s [E] for 15.0 minutes. What is the displacement of the train?

vave = d t

12.0 m/s [E] = d 900 s= 10800 m [E] = 10.8 km [E]

t = 15.0 min x 60 s/min = 900 s

• When we graph to demonstrate velocity we use a position time graph

time t(s) Position d(m) [E]

0.0 0.0

2.0 10.0

4.0 20.0

6.0 30.0

8.0 40.0

10.0 50.0

time (s)

Pos

ition

d (

m)

0 2 4 6 8 10

10

20

30

4

0

5

0

Position- Time Graph

Try the Following

According to the data below, what is the velocity of the car?

Solution calculating the average

velocity:Δ

dΔt

=dfinal – dinitial

tfinal – tinitial

=40.0 m – 10.0 m

8.0 s – 2.0 s

= +5.0 m/s

riserun

slope =

=

= 5.0 m/s [E]

0.0 to 2.0 s

2.0 to 4.0 s

4.0 to 6.0 s

6.0 to 8.0 s

8.0 to 10.0 s

Time interval (s)

Try the Following

Plot the following Data.What type of motion is this?

Solution

Uniform Motion!

1.3 Acceleration

Acceleration

• Is a change in velocity (speeding up or slowing down)

• The unites = m/s2

• Positive (+) Acceleration= velocity • Negative(-) Acceleration = velocity

(deceleration)

Positive Acceleration

• Positive (+) acceleration occurs two ways:

1. If direction is positive (+) and velocity is increasing

2) If direction is negative (-) and velocity is decreasing

+ direction

- direction

increasing velocity

+ direction

- direction

decreasing velocity

Position - Time Graphs

Time t (s)

Pos

itio

n (m

) [E

]

• Positive acceleration the slope is increasing

Velocity – Time

Time (s)

Vel

ocit

y (m

/s)

[E]

• positive acceleration the slope is increasing • the slope gives the acceleration

Negative Acceleration• Negative” (-) acceleration occurs in two ways:

1. If direction is positive (+) and velocity is decreasing

2) If direction is negative (-) and velocity is increasing

+ direction

- direction

decreasing velocity

+ direction

- direction

increasing velocity

Time t (s)

Pos

ition

(m

) [E

]

• negative acceleration because the slope is decreasing

Time (s)

Vel

ocity

(m

/s)

[E]

• Negative acceleration since the slope is decreasing

• The area under gives the distance traveled.

http://videos.howstuffworks.com/hsw/9610-physics-of-motion-acceleration-and-deceleration-video.htm

Uniform Motion Accelerated Motion

Time (s, h, etc)

Distance(m, km, etc)

d-t Graph

slope = slope of tangent =

Time (s, h, etc)

Distance(m, km, etc)

d-t Graph

speed instantaneous speed

Uniform Motion Accelerated Motion

Time (s, h, etc)

Velocity(m/s or km/h)

v-t Graph

area = slope =

Time (s, h, etc)

Velocity(m/s or km/h)

v-t Graph

distance accelerationarea = distance

Uniform Motion Accelerated Motion

Time (s, h, etc)

Acceleration(m/s2)

a-t Graphno a-t Graph

area = velocity

acceleration = change in velocity

change in time

a = v t

a = vf – vi

t

where: t = change in time in s or h

v = velocity in m/s or km/h

vi = initial velocity in m/s or km/h

vf = final velocity in m/s or km/h

a = acceleration in m/s2

Example 1Rudy falls out of an airplane and after 8.0 s he is travelling at 78.48 m/s. What is his acceleration?

vf = 78.48 m/s

t = 8.0 s a = vf - vi t = 78.48 m/s – 0 m/s

8.0 s = 9.8 m/s2

vi = 0 m/s

Example 2The initial speed of a bicycle is 8.0 m/s and it is moving for 6.0 s. If the final speed is 10.0 m/s, what is the acceleration of the bicycle?

t = 6.0 s

vi = 8.0 m/s a = vf – vi

t a = 10.0 m/s – 8.0 m/s

6.0 s a = 0.33 m/s2

vf = 10.0 m/s

a = ?

Try the Following

What is the acceleration of a car if its speed is increased uniformly from 40 m/s to 70 m/s in 3.0 s?

Solution

What is the acceleration of a car if its speed is increased uniformly from 40 m/s to 70 m/s in 3.0 s?

a = vf – vi

ta = (70m/s – 40m/s)

3.0sa = 10 m/s2

Example 3

Mike is traveling down Franklin Ave at 50 km/h. He sees Joslin standing at the bus stop and hits the brakes so he can pick her up. How long will it take for him to come to a stop if his acceleration is –5.0 m/s2?

vi = 50km/h x 10003600

= 13.88888 m/svf = 0 m/st = ?a = -5.0 m/s2

a = vf – vi

t-5.0 m/s2 = (0m/s – 13.8888 m/s)

t t = - 13.888888 m/s

-5.0m/s2

t = 2.8 s

Try the Following

Jack is traveling down Thickwood at 10 km/h. He sees Amber walking at the side of the road and hits the brakes so he can pick her up. How long will it take for him to come to a stop if his acceleration is –2.0 m/s2?

Solution

Jack is traveling down Thickwood at 10 km/h. He sees Amber walking at the side of the road and hits the brakes so he can pick her up. How long will it take for him to come to a stop if his acceleration is –2.0 m/s2?

vi = 10km/h x 10003600

= 2.7777 m/svf = 0 m/st = ?a = -2.0 m/s2

a = vf – vi

t-2.0 m/s2 = (0m/s – 2.7777 m/s)

t t = - 2.7777 m/s

-2.0m/s2

t = 1.38 s

Dan is driving down the highway and comes up behind a logging truck traveling 22 m/s. He hits the brakes, and accelerates uniformly at a rate of - 2.5 m/s2 for 5.0 seconds until he reaches the same speed as the truck. What was his initial speed in m/s and km/h?

Try the Following

Dan is driving down the highway and comes up behind a logging truck traveling 22 m/s. He hits the brakes, and accelerates uniformly at a rate of - 2.5 m/s2 for 5.0 seconds until he reaches the same speed as the truck. What was his initial speed in m/s and km/h?

vi = ? m/svf = 22 m/st = 5.0 sa = -2.5 m/s2

34.5m/s x 3600 1000

= 124 km/h = 1.2 x102 km/h

a = vf – vi

t-2.5 m/s2 = (22 m/s – x)

5.0 s(-2.5 m/s2)(5.0 s) = 22 m/s - x-12.5 m/s = 22 m/s - xx = 22m/s + 12.5 m/sx = 34.5 m/svi= 35 m/s

Solution

• Is 9.8 m/s2… (applies to all objects)

• It is greater near sea level

• It is less on the top of a mountain

• Larger masses have more

• There is always drag/air resistance… ignore it

Acceleration Due to Gravity

When Doing Calculations:

• if an object is falling:

• if an object is going up:

a = 9.81 m/s2

a = 9.81 m/s2

Example 1How fast are you falling after 2.5 s of free fall. Remember a = 9.81 m/s2

vi = 0 m/s

a = 9.81 m/s2

t = 2.5 s

vf = ?

vf = vi + at

vf = 0 m/s + (9.81 m/s2)(2.5 s)

vf = 24.525 m/s

vf = 25 m/s

Calculating average speed/velocity for constant acceleration

vave = vi + vf 2

vave= average speedvi = initial speedvf= final speed

Example 1 A train traveling through the Rocky Mountains, enters the Kicking Horse traveling at 35 m/s. When it reaches the top of the pass 65 minutes later it has slowed down to 15 m/s. What is the average speed of the train?

vave = vi + vf

2

Vave = 25 m/s

Vave = 35 m/s + 15 m/s2

Example 2 A car traveling is travelling up Thickwood at 40 m/s. When it reaches the top of the hill 3 minutes later it has slowed down to 10 m/s. What is the average speed of the car?

vave = vi + vf

2

Vave = 25 m/s

Vave = 40 m/s + 10 m/s2

Try A car traveling is travelling up highway 63 at 80 m/s. When it reaches the Fort McMurray 30 minutes later it has slowed down to 6 m/s. What is the average speed of the car?

Solution A car traveling is travelling up highway 63 at 80 m/s. When it reaches the Fort McMurray 30 minutes later it has slowed down to 6 m/s. What is the average speed of the car?

vave = vi + vf

2

Vave = 43 m/s

Vave = 80 m/s + 6 m/s2

Work and Energy

Force

• Is any push or pull on an object

• It is measured in Newtons (N)

• Objects remain at rest unless unbalanced forces act upon it

Balanced & Unbalanced Forces• Balanced force:

– forces are the same size but in the opposite direction

– cancel each other out.

• Unbalanced force– Forces are in the opposite

direction– one force is larger than the

other

• Deceleration (slowing down)• the force is in the opposite

direction of the movement

• energy is transferred from the source of the force to the object that the force is acting upon

• Accelerating (speeding up)• the force is in the same

direction as the moving object

To Change the Motion of Objects

• A force is needed

F = ma

F = force (Kg • m/s2) or 1 Newton (N)

m = mass (kg)

a = acceleration (m/s2)

• Note: weight is the force due to gravity ( 9.81 m/s2)

Example1 A 1000 kg car is accelerated at 2.5 m/s2. What is the force acting on it?

Example1 A 1000 kg car is accelerated at 2.5 m/s2. What is the force acting on it?

F = ma F = (1000kg)(2.5 m/s2 )F = 2500 N

F = 2.5 x 103 N

Solution

Try the Following

A 500 kg car is accelerated at 4.5 m/s2. What is the force acting on it?

A 500 kg car is accelerated at 4.5 m/s2. What is the force acting on it?

F = ma F = (500kg)(4.5 m/s2 )F = 2250 N

F = 2.2 x 103 N

Solution

Example 2 What is the mass of a crate with a weight of 450 N?

F = ma450 N = m 9.81 m/s2

m = 450 N/9.81 m/s2

m = 45.9 kg

F = 450 Nm = ?a = 9.81 m/s2

Example 3What force is needed to accelerate a 500 kg car from rest to 20 m/s in 5.0 s?

F = ? Nm = 500 kga = ?a = vf – vi

ta = (20m/s - 0 m/s)

5.0 sa = 4.0 m/s2

F = maF = (500 kg)(4.0 m/s2 )F = 2000 NF = 2.0 x 103 N

Work

• Occurs when a person lifts a weight, shovels snow or pushes a car

• Occurs when a force acts through a distance