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Science 10 Physics Unit B Read pg

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Conversions Review

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How to Multiply Fractions 3 x 4 = x 7 = 8 3 x 5 =

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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 m m

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E.g. 580 m km 1000 m = 1 km 580m x 1 km = 1000 m km

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Try Unit Conversions (Time) E.g 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 h s 1 hr = 3600 s 2.10 h x 3600 s = 7560 s 1h Try: 3 h s 3 h x 3600s = 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 second 2 (m/s 2 ) 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 v ave = v i + v f 2 a = v f - v i t F = ma W = Fd Ep = mgh Ek = 1/2 mv 2 % 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) Title Scale *Note: In Physics, time will always be the horizontal axis

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v ave = v i + v f 1. solve for v i 22. solve for v f a = v f - v i 3. solve for v i t 4. solve for t F = ma 5. solve for a W = Fd6. sove for F Ep = mgh 7. solve for h Ek = 1/2 mv 2 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 graph Time (s)Distance (m)

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Read pg

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

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Average Speed = distance traveled time v = d t v = d final – d initial t final – t initial 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 ________ ΔdΔd ΔtΔt

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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: ΔdΔd ΔtΔt v =

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b.) If Lance Armstrong bikes m in 10.0 s, what is the cyclists average speed? v = ΔdΔd ΔtΔt m 10.0 s 20.0 m/s d = m t = 10.0 s v = ? v =

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c.) If a train traveled 100 km in hours what is its speed in km/h and in m/s? d = 100 km t = h v = ? 100 km h 200 km/h ΔdΔd ΔtΔt 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/s t = ? t = d v t = 1000m 15 m/s v = t = 67 s ________ ΔdΔd ΔtΔt

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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 = ΔdΔd ΔtΔt

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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 = ΔdΔd ΔtΔt

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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 = ΔdΔd ΔtΔt

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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 h v = 95.5 km/h = /3.6 = 26.5 m/s ________ ΔdΔd ΔtΔt 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 = ΔdΔd ΔtΔt

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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) distance (m) time (s) d t Distance-Time Graph

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The steepness of the graph is the slope Example: slope = y 2 – y 1 x 2 – x 1 = 80m – 20m 4.0 s- 1.0 s = 20 m/s slope = rise run = y 2 – y 1 x 2 – x 1

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The steeper the slope the higher the speed. Distance (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 = = ΔdΔd ΔtΔt 30 m – 10 m 6.0 s – 2.0 s = 5.0 m/s V

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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 motion Upward slope = speed is increasing Downward slope = speed is decreasing

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Time t (s) Speed v (m/s) Time (s) Speed (m/s) Uniform Motion

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Can be used to determine the distance an object travels…… calculate the area under the line Example 1 Calculate the area under the following speed-time graph up to 10.0 s. Time (s) Speed (m/s)

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Solution Calculate 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) Speed (m/s)

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Example 2 Calculate distance travelled by an object in 20.0s. Time (s) Speed (m/s)

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Solution Calculate distance travelled by an object in 20.0s. Time (s) Speed (m/s) 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) Speed (m/s) Try the Following

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Calculate distance travelled by an object in 40.0s. Time (s) Speed (m/s) Solution A= 1 b x h 2 A= 1 40 x 25 2 A = 500 m

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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 -Degreesx or y axis -Degreescompass 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 E W 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 axis A B 10 km 1.0 m/s vector A: vector B: 10 km, 80 E of the y axis 10 km, 80 E of S 1.0 m/s, 26 S of the x axis 1.0 m/s, 26 S of W

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N E S W 30º 15º 55º 30° S of E 15°N of W 55°W of S S y axis + y axis + x axis x axis A B 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 m 40 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 direction 20 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 (d final – d initial ) t = time

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Velocity ( ) Describes the rate of motion and the direction of the objects motion It is a vector quantity the formula is: v ave = d t v ave = d final – d initial t final - t initial v

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Example 3 A 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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Solution A 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 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] 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 = …. h 558 km/h x 1000 = 155 m/s [S] 3600 v ave = d t = 465 km [S] – 0 km … h = 558 km/h [S]

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Example 5 A train travels at 12.0 m/s [E] for 15.0 minutes. What is the displacement of the train? v ave = d t 12.0 m/s [E] = d 900 s = 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] time (s) Position d (m) 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: calculating the average velocity: ΔdΔd ΔtΔt = d final – d initial t final – t initial = 40.0 m – 10.0 m 8.0 s – 2.0 s = +5.0 m/s rise run 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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Acceleration Is a change in velocity (speeding up or slowing down) The unites = m/s 2 Positive (+) Acceleration= velocity Negative(-) Acceleration = velocity (deceleration)

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Positive Acceleration Positive (+) acceleration occurs two ways: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) Position (m) [E] Positive acceleration the slope is increasing

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Velocity – Time Time (s) Velocity (m/s) [E] positive acceleration the slope is increasing the slope gives the acceleration

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Negative Acceleration N egative (-) 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) Position (m) [E] negative acceleration because the slope is decreasing

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Time (s) Velocity (m/s) [E] Negative acceleration since the slope is decreasing The area under gives the distance traveled.

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deceleration-video.htm

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Uniform MotionAccelerated 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 MotionAccelerated 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 distanceacceleration area = distance

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Uniform MotionAccelerated Motion Time (s, h, etc) Acceleration (m/s 2 ) a-t Graph no a-t Graph area = velocity

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acceleration = change in velocity change in time a = v t a = v f – v i t where: t = change in time in s or h v = velocity in m/s or km/h v i = initial velocity in m/s or km/h v f = final velocity in m/s or km/h a = acceleration in m/s 2

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Example 1 Rudy falls out of an airplane and after 8.0 s he is travelling at m/s. What is his acceleration? v f = m/s t = 8.0 s a = v f - v i t = m/s – 0 m/s 8.0 s = 9.8 m/s 2 v i = 0 m/s

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Example 2 The 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 v i = 8.0 m/s a = v f – v i t a = 10.0 m/s – 8.0 m/s 6.0 s a = 0.33 m/s 2 v f = 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 = v f – v i t a = (70m/s – 40m/s) 3.0s a = 10 m/s 2

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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/s 2 ? v i = 50km/h x = m/s v f = 0 m/s t = ? a = -5.0 m/s 2 a = v f – v i t -5.0 m/s 2 = (0m/s – m/s) t t = m/s -5.0m/s 2 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/s 2 ?

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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/s 2 ? v i = 10km/h x = m/s v f = 0 m/s t = ? a = -2.0 m/s 2 a = v f – v i t -2.0 m/s 2 = (0m/s – m/s) t t = m/s -2.0m/s 2 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 m/s 2 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 m/s 2 for 5.0 seconds until he reaches the same speed as the truck. What was his initial speed in m/s and km/h? v i = ? m/s v f = 22 m/s t = 5.0 s a = -2.5 m/s m/s x = 124 km/h = 1.2 x10 2 km/h a = v f – v i t -2.5 m/s 2 = (22 m/s – x) 5.0 s (-2.5 m/s 2 )(5.0 s) = 22 m/s - x m/s = 22 m/s - x x = 22m/s m/s x = 34.5 m/s v i = 35 m/s Solution

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Is 9.8 m/s 2 … (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 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/s 2

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Example 1 How fast are you falling after 2.5 s of free fall. Remember a = 9.81 m/s 2 v i = 0 m/s a = 9.81 m/s 2 t = 2.5 s v f = ? v f = v i + at v f = 0 m/s + (9.81 m/s 2 )(2.5 s) v f = m/s v f = 25 m/s

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Calculating average speed/velocity for constant acceleration v ave = vi + vf 2 v ave = average speed vi = initial speed vf= 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? v ave = v i + v f 2 Vave = 25 m/s Vave = 35 m/s + 15 m/s 2

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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? v ave = v i + v f 2 Vave = 25 m/s Vave = 40 m/s + 10 m/s 2

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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? v ave = v i + v f 2 Vave = 43 m/s Vave = 80 m/s + 6 m/s 2

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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/s 2 ) or 1 Newton (N) m = mass (kg) a = acceleration (m/s 2 ) Note: weight is the force due to gravity ( 9.81 m/s 2 )

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Example1 A 1000 kg car is accelerated at 2.5 m/s 2. What is the force acting on it?

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Example1 A 1000 kg car is accelerated at 2.5 m/s 2. What is the force acting on it? F = ma F = (1000kg)(2.5 m/s 2 ) F = 2500 N F = 2.5 x 10 3 N Solution

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Try the Following A 500 kg car is accelerated at 4.5 m/s 2. What is the force acting on it?

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F = ma F = (500kg)(4.5 m/s 2 ) F = 2250 N F = 2.2 x 10 3 N Solution

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Example 2 What is the mass of a crate with a weight of 450 N? F = ma 450 N = m 9.81 m/s 2 m = 450 N/9.81 m/s 2 m = 45.9 kg F = 450 N m = ? a = 9.81 m/s 2

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Example 3 What force is needed to accelerate a 500 kg car from rest to 20 m/s in 5.0 s? F = ? N m = 500 kg a = ? a = v f – v i t a = (20m/s - 0 m/s) 5.0 s a = 4.0 m/s 2 F = ma F = (500 kg)(4.0 m/s 2 ) F = 2000 N F = 2.0 x 10 3 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

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