science 10 physics unit b read pg. 465 - 477. conversions review
TRANSCRIPT
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Science 10 Physics
Unit BRead pg. 465 - 477
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Conversions Review
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How to Multiply Fractions
• 3 x 4 = 1 5
• 3 x 7 = 8
• 3 x 5 = 9
12 5
21 8
15 9
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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
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• 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
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• 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
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• E.g. 580 m km 1000 m = 1 km
580m x 1 km = 1000 m
0.580 km
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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
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• 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
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• 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 (%)
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• 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
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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
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Drawing Graphs
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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)
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Example:
Distance-Time Graph
Label (Units)
Label (Units)
Title
Scale
*Note: In Physics, time will always be the horizontal axis
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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:
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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
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1.1 MotionRead pg.126-131
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Energy
• Causes changes in the motion to occur to an object
• It can speed objects up, slow them down or change their direction
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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
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v = d t
= changed = distance in m or kmt = time in s or hv = speed in m/s or km/h
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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)
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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
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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 =
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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 =
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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 =
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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
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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
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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
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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
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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
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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
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• 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
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• speed describes the rate of motion an object has
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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
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• 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
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• 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
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Try the Following:
• Make a Distance time graph for the following
• Calculate the speed of the boat
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v = =Δd
Δt30 m – 10 m
6.0 s – 2.0 s
= 5.0 m/sV
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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
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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
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• 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
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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
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Example 2Calculate distance travelled by an object in 20.0s.
Time (s)
Spe
ed (
m/s
)
20.0
10.0
0.0
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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
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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
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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
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1.2 Velocity
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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]
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• 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)
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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
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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
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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
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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
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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)
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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
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Example1
A person runs 25 m south and then another 15 m south.
(a) What is the distance travelled?
(b) What is the displacement?
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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
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• 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
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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?
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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
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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
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• 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
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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
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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] =
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d) velocity v = d/t = 2.0 m [N]/5.00 s = 0.40 m/s [N]
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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]
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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
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• 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
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Try the Following
According to the data below, what is the velocity of the car?
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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]
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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?
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Solution
Uniform Motion!
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1.3 Acceleration
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Acceleration
• Is a change in velocity (speeding up or slowing down)
• The unites = m/s2
• Positive (+) Acceleration= velocity • Negative(-) Acceleration = velocity
(deceleration)
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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
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Position - Time Graphs
Time t (s)
Pos
itio
n (m
) [E
]
• Positive acceleration the slope is increasing
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Velocity – Time
Time (s)
Vel
ocit
y (m
/s)
[E]
• positive acceleration the slope is increasing • the slope gives the acceleration
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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
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Time t (s)
Pos
ition
(m
) [E
]
• negative acceleration because the slope is decreasing
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Time (s)
Vel
ocity
(m
/s)
[E]
• Negative acceleration since the slope is decreasing
• The area under gives the distance traveled.
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http://videos.howstuffworks.com/hsw/9610-physics-of-motion-acceleration-and-deceleration-video.htm
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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
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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
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Uniform Motion Accelerated Motion
Time (s, h, etc)
Acceleration(m/s2)
a-t Graphno a-t Graph
area = velocity
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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
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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
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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 = ?
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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?
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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
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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
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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?
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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
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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
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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
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• 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
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When Doing Calculations:
• if an object is falling:
• if an object is going up:
a = 9.81 m/s2
a = 9.81 m/s2
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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
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Calculating average speed/velocity for constant acceleration
vave = vi + vf 2
vave= average speedvi = initial speedvf= final speed
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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
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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
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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?
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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
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Work and Energy
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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
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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
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• 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
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To Change the Motion of Objects
• A force is needed
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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)
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Example1 A 1000 kg car is accelerated at 2.5 m/s2. What is the force acting on it?
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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
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Try the Following
A 500 kg car is accelerated at 4.5 m/s2. What is the force acting on it?
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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
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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
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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
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Work
• Occurs when a person lifts a weight, shovels snow or pushes a car
• Occurs when a force acts through a distance