3 How to Multiply Fractions 3 x =3 x =83 x =9125218159
4 Unit Conversions (Distance) E.g km m1 km = 1000 mTry: m m3750 m3.75 km x m =1 km5850 m5.85 km x m =1 km
5 E.g. 427cm m100 cm = 1 mTry: cm m427cm x 1 m =100cm4.27 m8.64 cm865 cm x 1m =100cm
6 E.g. 67 mm m1000 mm = 1 mTry: mm m0.067 m67 mm x 1 m =1000 mm0.765 m765 mm x 1m =1000 mm
7 E.g m km1000 m = 1 km0.580 km580m x 1 km =1000 m
8 Try Unit Conversions (Time) E.g h min1 hr = 60 min2.75 h x 60 min = 165 min1 hTry:42 min h42 min x 1h = h60 min8
9 E.g h s1 hr = 3600 s2.10 h x s = 7560 s1hTry:3 h s3 h x s = s1 h
10 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 (%)
11 Formulas to know for this unit: v = d t vave = vi + vf2a = vf - vitF = maW = FdEp = mghEk = 1/2 mv2% Efficiency = useful output x 100%total input11
12 Rearranging formulasYou need to ISOLATE the variable you are trying to solve forWhat ever mathematical operation you do to one side of the = you need to also do to the other sideEx. v = d Solve for d and solve for tt
14 Parts of a Graph All graphs should have: A horizontal axis (or x axis, ALWAYS TIME!)A vertical axis (or y axis)A titleLabels on each axisUnits for each axisAppropriate scale (numbering on both axis)
15 Example: Title Distance-Time Graph Label (Units) *Note: In Physics, time will always be the horizontal axisScaleLabel (Units)
16 Rearrange the following formulas: vave = vi + vf 1. solve for vi2 2. solve for vfa = vf - vi 3. solve for vit solve for tF = ma solve for aW = Fd 6. sove for FEp = mgh solve for hEk = 1/2 mv2 8. solve for v% Eff = useful output x 100% 9. solve for total input total input solve for total output
17 Draw the following graph Time (s)Distance (m)0.02.05.04.010.06.015.08.020.025.012.030.014.035.016.040.0
19 Energy Causes changes in the motion to occur to an object It can speed objects up, slow them down or change their direction
20 Uniform Motion Describes a type of movement It occurs when an object travels in a straight line at a constant speedis difficult to maintain so ….we use AVERAGE SPEED
21 t = change d = distance in m or km t = time in s or h v = dt = changed = distance in m or kmt = time in s or hv = speed in m/s or km/h
22 Average Speed = distance traveled time v = d t v = dfinal – dinitialtfinal – tinitialv = speed (m/s or km/h)d = change in distance (m or km)t = change in time (s or h)
23 Example 1A baseball travels 200 m in seconds. What is the average speed of the baseball?________________ΔdΔtv =d = 200 mv = ?t = 1.50 sv = 200m1.50 sv = 133 m/s
24 Try:a.) A baseball travels 20.0 m in seconds. Calculate the average speed.d = 20.0 mΔdΔtv =t = 1.50 sv = ?v = 20.0 m1.50 sv = 13.3 m/s
25 b. ). If Lance Armstrong bikes 200. 0 m in 10. 0 s, b.) If Lance Armstrong bikes m in 10.0 s, what is the cyclist’s average speed?ΔdΔtv =d = mt = 10.0 s200.0 m10.0 sv =v = ?v =20.0 m/s
26 c. ). If a train traveled 100 km in 0. 500 hours c.) If a train traveled 100 km in hours what is its speed in km/h and in m/s?ΔdΔtd = 100 kmv =100 km0.500 h200 km/ht = hv =v = ?v =200 km x 1000 m/km1 h x 60 min/h x 60 sec/min=200 km/h3.6= 55.6 m/s
27 Example 2A car travels 1.00 km at a constant speed of 15 m/s. What time is required to cover this distance?________________ΔdΔtv =d = 1.00 km= 1000 mv = 15 m/st = ?t = dvt = m15 m/st = 67 s
28 Try:a.) How long would it take a car to travel m if its speed was 40.0 m/s?ΔdΔtd = 4000 mv =v = 40.0 m/st = dv= 4000 m40.0 m/s= 100 st = ?
29 b. ). How long would it take a car to travel b.) How long would it take a car to travel m if its speed was 10.0 m/s?ΔdΔtd = 2000 mv =v = 10.0 m/st = dv= 2000 m10.0 m/s= 200 st = ?
30 c. ). How long would it take a car to travel c.) How long would it take a car to travel m if its speed was 30.0 m/s?ΔdΔtd = 8000 mv =v = 30.0 m/st = dv= 8000 m30.0 m/s= 267 st = ?
31 Example 3A motorist travels 406 km in 4 hours and 15 minutes. What is the average speed in km/h and m/s?________________ΔdΔtv =v = 406 km4.25 hv = km/hd = 406 kmv= ?t = 4 hour+ 15min60 min/h= 4.25 h= /3.6= 26.5 m/sv
32 Example 4. How far of a distance will a car cover if it travels 2 Example 4. How far of a distance will a car cover if it travels 2.00 m/s for 1.00 min?ΔdΔtd = ?v =v = 2.00 m/sd = v t=2.00 x 60.0= 120 mt = 1.00 min= 60.0 s32
33 Distance Time Graphsdistance varies directly with time when speed is constantHave the following components:time is the (x-axis)distance is the (y-axis)the slope of the line is the speed of an object
34 speed describes the rate of motion an object has
35 Distance-Time Graph d t t(s) d(m) 1.0 20 2.0 40 3.0 60 4.0 80 5.0 1.0202.0403.0604.0805.0100Distance-Time Graphdistance (m) d ttime (s)
36 The steepness of the graph is the slope Example:slope = y2 – y1x2 – x1= 80m – 20m4.0 s- 1.0 s= 20 m/sslope = riserun= y2 – y1x2 – x1
37 The steeper the slope the higher the speed. ABDistance (m)Ctime (s)Which line shows the greatest speed? The slowest speed?
38 Try the Following: Make a Distance time graph for the following Calculate the speed of the boat
40 Speed- Time GraphsThe area under the graph is the distance an object travelsThe slope of the line gives you information about the speedE.g.A slope of zero (flat line) = uniform motionUpward slope = speed is increasingDownward slope = speed is decreasing
41 Time t (s) Speed v (m/s) 0.0 5.00 2.0 4.0 6.0 8.0 10.0 Uniform Motion Speed (m/s)Uniform MotionTime (s)
42 Calculate the area under the following speed-time graph up to 10.0 s. Can be used to determine the distance an object travels…… calculate the area under the lineExample 1Calculate the area under the following speed-time graph up to 10.0 s.Time (s)Speed (m/s)10.05.010.00.05.0
43 Calculate the area under the following speed-time graph up to 10.0 s. SolutionCalculate the area under the following speed-time graph up to 10.0 s.Time (s)Speed (m/s)10.05.010.0area = wA = (10.0 s)(5.0 m/s)A = 50.0 m0.0
44 Calculate distance travelled by an object in 20.0s. Example 2Calculate distance travelled by an object in 20.0s.Time (s)Speed (m/s)20.010.00.0
45 Calculate distance travelled by an object in 20.0s. SolutionCalculate distance travelled by an object in 20.0s.Time (s)Speed (m/s)20.010.0A= 1 b x h2A= x 10A = 100 m0.0
46 Try the Following Calculate distance travelled by an object in 40.0s. 30.0Time (s)Speed (m/s)20.010.020.00.040.0
47 Solution Calculate distance travelled by an object in 40.0s. 30.0Time (s)Speed (m/s)20.010.020.0A= 1 b x h2A= x 25A = 500 m0.040.0
49 Vectors Verses Scalar Scalar quantities: Vector quantities: involve only magnitude (amount)Vector quantities:involve both magnitude and directionAre drawn using arrowsE.g.) Speed = 20 m/sE.g.) Displacement = 20 m [N]Velocity = 20 m/s [N]
50 when describing a vector, we have two quantities that indicate the direction: from the-Degreesx or y axisusing-Degreescompass directions(N, S, E, or W)
51 Cartesian Method E.g. 6 m [30°] E.g. 10 m [right] up (90°) left (180°) down (270°)right (0°)left (180°)30°6 m10 mE.g. 6 m [30°]40°2 m8 mE.g. 10 m [right]
52 (+) (+) (-) (-) Navigator Method (+) (+) (-) (-) N W E S angle (°)N of Wangle (°)N of E(+)It uses N-S-E-W .. the angle is relative to E (east) or W (west), or a direction at N, S, E or WWE(-)angle (°)S of Wangle (°)S of E(-)(-)(-)S
53 N E S + y axis y axis + x axis x axis vector A: vector B: W+ y axis y axis+ x axis x axisAB10 km1.0 m/s8065vector A:vector B:10 km, 80 E of the y axis10 km, 80 E of S1.0 m/s, 26 S of the x axis1.0 m/s, 26 S of W
54 + y axis N 15°N of W x axis W E + x axis 30° S of E 55°W of S S C15°N of W15º x axisWE+ x axis30º55º30° S of E55°W of SABSS y axis
55 Distance Verses Displacement Distance (d)is how far an object travelsIt is a scalar quantity (magnitude)Displacement (d)Is change in both distance and directionIt is a vector quantity (magnitude and direction)
56 Vectors Direction Vectors in the Same Direction: 10 m 5 m = 15 m To find the distance:Add them togetherTo find the displacement:Add them together and include the direction10 m5 m= 15 m
57 Example1A person runs 25 m south and then another 15 m south.(a) What is the distance travelled? (b) What is the displacement?
58 SolutionA person runs 25 m south and then another 15 m south.(a) What is the distance travelled? (b) What is the displacement?25 m40 m15 m40 m South
59 Vectors in Opposite Directions: To find the distance: Add them togetherTo find the displacement: take the difference between the two numbers and include the direction20 mDistance = 25 mDisplacement = 15 m E5 m
60 a) What is the distance travelled? Example 2A 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?
61 a) What is the distance travelled? SolutionA 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 km150 km200km50 km50 km N
62 Speed verses Velocity Speed (v): is the rate of an objects motion It is a scalar quantityThe formula is:v= d tv = speed d = distance (dfinal – dinitial) t = time
63 Velocity ( )Describes the rate of motion and the direction of the object’s motionIt is a vector quantitythe formula is:vave = d tvave = dfinal – dinitialtfinal - tinitialv
64 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) distanceb) displacementc) speedd) velocity
65 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) distanced = 35.0 m m =68.0 mb) displacementd = 35.0 m [N] m [S] =2.0 m [N]c) speed v = d/t= 68.0 m/5.00 s= 13.6 m/s
66 d) velocity v = d/t= 2.0 m [N]/5.00 s= 0.40 m/s [N]
67 Example 4A 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?vave = d t= 465 km [S] – 0 km0.8333… h= 558 km/h [S]t = min60min/h= …. h558 km/h x = 155 m/s [S]3600
68 Example 5A train travels at 12.0 m/s [E] for 15.0 minutes. What is the displacement of the train?vave = d t12.0 m/s [E] = d900 s= m [E]= 10.8 km [E]t = 15.0 min x 60 s/min= 900 s
69 When we graph to demonstrate velocity we use a position time graph time t(s)Position d(m) [E]0.02.010.04.020.06.030.08.040.050.0Position d (m)time (s)
70 Try the FollowingAccording to the data below, what is the velocity of the car?
71 Solution rise Δd slope = = run Δt calculating the average velocity: =dfinal – dinitialtfinal – tinitialriserunslope ===40.0 m – 10.0 m8.0 s – 2.0 s= +5.0 m/s= 5.0 m/s [E]
72 Try the Following Plot the following Data. 0.0 to 2.0 s2.0 to 4.0 s4.0 to 6.0 s6.0 to 8.0 s8.0 to 10.0 sTime interval (s)Plot the following Data.What type of motion is this?
75 Acceleration Is a change in velocity (speeding up or slowing down) The unites = m/s2Positive (+) Acceleration= velocityNegative(-) Acceleration = velocity (deceleration)
76 Positive Acceleration Positive (+) acceleration occurs two ways:If direction is positive (+) and velocity is increasing+ direction- directionincreasing velocityIf direction is negative (-) and velocity is decreasing+ direction- directiondecreasing velocity
77 Position - Time Graphs Position (m) [E] Time t (s) Positive acceleration the slope is increasing
78 Velocity – Time Velocity (m/s) [E] Time (s) positive acceleration the slope is increasingthe slope gives the acceleration
79 Negative Acceleration Negative” (-) acceleration occurs in two ways:If direction is positive (+) and velocity is decreasing+ direction- directiondecreasing velocityIf direction is negative (-) and velocity is increasing+ direction- directionincreasing velocity
80 Position (m) [E] Time t (s) negative acceleration because the slope is decreasing
81 Time (s) Velocity (m/s) [E] Negative acceleration since the slope is decreasingThe area under gives the distance traveled.
86 acceleration = change in velocity change in time a = vf – vita = vtwhere: t = change in time in s or hv = velocity in m/s or km/hvi = initial velocity in m/s or km/hvf = final velocity in m/s or km/ha = acceleration in m/s2
87 Example 1Rudy falls out of an airplane and after 8.0 s he is travelling at m/s. What is his acceleration?a = vf - vit= m/s – 0 m/s8.0 s= 9.8 m/s2t = 8.0 svf = m/svi = 0 m/s
88 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?vi = 8.0 m/sa = vf – vita = 10.0 m/s – 8.0 m/s6.0 sa = 0.33 m/s2t = 6.0 svf = 10.0 m/sa = ?
89 Try the FollowingWhat is the acceleration of a car if its speed is increased uniformly from 40 m/s to 70 m/s in 3.0 s?
90 SolutionWhat 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
91 Example 3Mike 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= m/svf = 0 m/st = ?a = -5.0 m/s2a = vf – vi t-5.0 m/s2 = (0m/s – m/s)tt = m/s-5.0m/s2t = 2.8 s
92 Try the FollowingJack 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?
93 SolutionJack 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= m/svf = 0 m/st = ?a = -2.0 m/s2a = vf – vi t-2.0 m/s2 = (0m/s – m/s)tt = m/s-2.0m/s2t = 1.38 s
94 Try the FollowingDan 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/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?
95 Solution vi = ? m/s vf = 22 m/s t = 5.0 s a = -2.5 m/s2 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/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?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 m/sx = 34.5 m/svi= 35 m/svi = ? m/svf = 22 m/st = 5.0 sa = -2.5 m/s234.5m/s x 36001000= 124 km/h = 1.2 x102 km/h
96 Acceleration Due to Gravity Is 9.8 m/s2… (applies to all objects)It is greater near sea levelIt is less on the top of a mountainLarger masses have moreThere is always drag/air resistance… ignore it
97 When Doing Calculations: if an object is falling:if an object is going up:a = 9.81 m/s2a = 9.81 m/s2
98 Example 1How fast are you falling after 2.5 s of free fall. Remember a = 9.81 m/s2vi = 0 m/sa = 9.81 m/s2t = 2.5 svf = ?vf = vi + atvf = 0 m/s + (9.81 m/s2)(2.5 s)vf = m/svf = 25 m/s
99 Calculating average speed/velocity for constant acceleration vave = vi + vf2vave= average speedvi = initial speedvf= final speed99
100 Example 1A 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 + vf2Vave = 35 m/s + 15 m/s2Vave = 25 m/s
101 Example 2A 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 + vf2Vave = 40 m/s + 10 m/s2Vave = 25 m/s
102 TryA 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?
103 SolutionA 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 + vf2Vave = 80 m/s + 6 m/s2Vave = 43 m/s
105 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
106 Balanced & Unbalanced Forces forces are the same size but in the opposite directioncancel each other out.Unbalanced forceForces are in the opposite directionone force is larger than the other
107 Deceleration (slowing down) the force is in the opposite direction of the movementenergy is transferred from the source of the force to the object that the force is acting uponAccelerating (speeding up)the force is in the same direction as the moving object
108 To Change the Motion of Objects A force is needed
109 F = force (Kg • m/s2) or 1 Newton (N) m = mass (kg) F = maF = 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)
110 Example1A 1000 kg car is accelerated at 2.5 m/s2. What is the force acting on it?
111 SolutionExample1A 1000 kg car is accelerated at 2.5 m/s2. What is the force acting on it?F = maF = (1000kg)(2.5 m/s2 )F = 2500 NF = 2.5 x 103 N
112 Try the FollowingA 500 kg car is accelerated at m/s2. What is the force acting on it?
113 SolutionA 500 kg car is accelerated at 4.5 m/s2. What is the force acting on it?F = maF = (500kg)(4.5 m/s2 )F = 2250 NF = 2.2 x 103 N
114 What is the mass of a crate with a weight of 450 N? Example 2What is the mass of a crate with a weightof 450 N?F = ma450 N = m 9.81 m/s2m = 450 N/9.81 m/s2m = 45.9 kgF = 450 Nm = ?a = 9.81 m/s2
115 Example 3What force is needed to accelerate a 500 kg car from rest to 20 m/s in 5.0 s?F = maF = (500 kg)(4.0 m/s2 )F = 2000 NF = 2.0 x 103 NF = ? Nm = 500 kga = ?a = vf – vita = (20m/s - 0 m/s)5.0 sa = 4.0 m/s2
116 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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