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Motion in a Straight Line

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KINEMATICS - the process of motion is integral to the description of matter characteristics - all matter is moving - therefore a method must be formulated for accuracy

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DISTANCE vs. DISPLACEMENT 1. DISTANCE - defined as the magnitude or length of motion - NO DIRECTION INDICATED symbol = d 2. DISPLACEMENT - magnitude and direction of motion symbol = s

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Measurement of Speed Total distance : the sum of the all changes in position Time : an interval of change measured in seconds Position : Separation between the object and the reference point

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Types of Speed Rest : no change in motion Instantaneous: the current speed of an object at a point of time Average : two ways to determine the “mean” movement Total distance / Total time Sum of the individual speeds / number of speed measurements Speed is a scalar quantity

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Velocity Displacement : the change in position based on distance and Direction Velocity : is the change in Distance per unit time with a specific direction Velocity is a vector quantity Velocity also has the following conditions: Average Instantaneous

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Constant Speed An object could be moving at a steady rate. Thus, its average and instantaneous speed would be the same! GBS Physics - speed vs. velocity

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Question: A car is traveling at a constant 60mi/hr in a circular path. Does it have a constant speed? Does it have a constant velocity?

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Problem Solving G – write down what information is given (with units) U – Identify what you have to find – the unknown ( with units ) E – Identify an equation (relationship) that equates the givens and unknown S – Solve the equation for the unknown (algebraically before substituting any of the "givens" with their units because there will be fewer mistakes in copying the units until the last step). S – Substitute the givens ( with units ) and find the answer.

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Sample Problem #1 If a car is traveling at an average speed of 60 kilometers per hour, how long does it take to travel 12 kilometers? GIVEN: Ave Speed = 60 km/hr Distance = 12 km UNKNOWN: Time - ? EQUATION : v = d/t

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Sample Problem #1 If a car is traveling at an average speed of 60 kilometers per hour, how long does it take to travel 12 kilometers? SOLVE : v = d/t => t = d/v SUBSTITUTION: t = 12km / 60km/h t =.2 h t = 12 min t = 720 sec

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Sample Problem #2 A high speed train travels 454 km in 7200 seconds. What is the train’s average speed? GIVEN: Time = 7200 s (always a good idea to convert to (base / fundamental units) - time to seconds- if needed!) Distance = 454 km - (always a good idea to convert to (base / fundamental units) - km to meters- if needed!) 454 km = 454000 m UNKNOWN: Ave Speed - ? EQUATION : v = d/t

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Sample Problem #2 A high speed train travels 454 km in 7200 seconds. What is the train’s average speed? SOLVE : v = d/t => v = d/t SUBSTITUTION: v = 454km / 7200s (with units!) v = 454000m / 7200s v = 63 m/s

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Example: If sun light takes about 8 minutes to go from the sun to the earth, how far away from the sun is the earth? Hint: light travels at 186,000 miles per second!!! GIVEN: Time = 8 min (always a good idea to convert to (base / fundamental units) - time to seconds- if needed!) 8 min = 480 s Ave Speed = 186,000 mi/s UNKNOWN: Distance - ? EQUATION : v = d/t

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Example: If sun light takes about 8 minutes to go from the sun to the earth, how far away from the sun is the earth? Hint: light travels at 186,000 miles per second!!! SOLVE : v = d/t => d = vt SUBSTITUTION: d = 186,000 mi/s X 480s (with units!) d = 89,000,000mi

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Summary Speed is based on position change relative to the origin Scalar and Vectors quantities are used to describe motion Calculations must include formula, substitution of proper units, and final solution in the MKS system.

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Relativity of Velocity Theory, developed in the early 20th century, which originally attempted to account for certain anomalies in the concept of relative motion, but which in its ramifications has developed into one of the most important basic concepts in physical science Velocity changes when compared to a frame of reference

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Acceleration Acceleration: rate of change of velocity Acceleration describes how fast an objects speed is changing per amount of time.

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Types of Acceleration Acceleration, also known as linear acceleration, rate at which the velocity of an object changes per unit of time. A = v/t ( Average Acceleration) Uniform Acceleration : the constant rate of change in Velocity ( Free Fall ) 9.81 m/s 2 (use 10 m/s 2 in multiple choice)

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If an object has a constant velocity, then its acceleration would be zero. If an object is slowing down, it is decelerating: or a NEGATIVE acceleration

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Formulas

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How to choose the best formula Free Fall Acceleration due to gravity Uniform acceleration Distance is not part of the question Time is part of the question

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How to choose the best formula Free Fall Acceleration due to gravity Uniform acceleration Distance is part of the question Time is part of the question

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How to choose the best formula Choose this formula when the question does not include the TIME

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HW p. #2– Q1 - a A 60 mi/hr wind is blowing toward the S. What is the resultant velocity of an airplane traveling 100 mi/hr when it is heading: Wind - 60 mi/hr S Plane - 100 mi/hr N Resultant = 40 mi/hr N

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HW p. #2– Q1 - b A 60 mi/hr wind is blowing toward the S. What is the resultant velocity of an airplane traveling 100 mi/hr when it is heading: Wind - 60 mi/hr S Plane - 100 mi/hr S Resultant = 160 mi/hr S

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HW p. #2– Q1 - c A 60 mi/hr wind is blowing toward the S. What is the resultant velocity of an airplane traveling 100 mi/hr when it is heading: Wind - 60 mi/hr S Plane - 100 mi/hr S Resultant = 116.6 mi/hr @ S 59º E a 2 + b 2 = c 2 (60 mi/hr) 2 + (100 mi/hr) 2 = c 2 C = 116.6 mi/hr = tan -1 (100 mi/hr / 60 mi/hr) = 59º

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HW p. #2– Q1 - c A 60 mi/hr wind is blowing toward the S. What is the resultant velocity of an airplane traveling 100 mi/hr when it is heading: GIVEN: a. v = 40 mi/hr; b. v = 160 mi/hr; c. v = 116.6 mi/hr ; t= 5 hr UNKNOWN: d - ? EQUATION : v = d/t SOLVE : v = d/t => d = vt SUBSTITUTION: a. d = (40 mi/hr)(5 hr) = 200 mi (with units!) b. d = (160 mi/hr)(5 hr) = 800 mi c. d = (116.6 mi/hr)(5 hr) = 583 mi

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HW p. #2– Q2 Rowboat across a stream flowing @ 3 mi/hr. Boy can row boat @ 4 mi/hr directly across stream. Water 3 mi/hr Boat - 4 mi/hr Resultant = 5 mi/hr @ 36.9º a 2 + b 2 = c 2 (3 mi/hr) 2 + (4 mi/hr) 2 = c 2 C = 5 mi/hr = tan -1 (3 mi/hr / 4 mi/hr) = 36.9º

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HW p. #3 – Q1 A fish swims at the rate of 2 ft/s. How long will it take this fish to swim 36 ft? GIVEN: Ave speed = 2 ft/s; d = 36 ft UNKNOWN: time - ? EQUATION : v = d/t SOLVE : v = d/t => t = d/v SUBSTITUTION: t = 36 ft / 2 ft/s (with units!) t = 18 s

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HW p. #3 – Q3 part a A car starts from rest & accelerates up to a velocity of 40 ft/s in 10 s? GIVEN: v i = 0 ft/s; v f = 40 ft/s; t = 10 s UNKNOWN: a - ? EQUATION : v f = v i + at SOLVE : v f = v i + at => v f - v i = at v f – v i / t = a SUBSTITUTION: a = v f – v i / t (with units!) a = ( 40 ft/s – O ft/s) / 10 s = 4 ft/s 2

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HW p. #3 – Q3 part b A car starts from rest & accelerates up to a velocity of 40 ft/s in 10 s? GIVEN: v i = 0 ft/s; v f = 40 ft/s; t = 10 s UNKNOWN: Ave speed = ? EQUATION : Ave speed = v f + v i / 2 SOLVE : ave V = v f + v i / 2 SUBSTITUTION: ave V = v f + v i /2 ave V = (40 ft/s + 0 ft/s) / 2 = 20 ft/s

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HW p. #3 – Q3 part c A car starts from rest & accelerates up to a velocity of 40 ft/s in 10 s? GIVEN: v i = 0 ft/s; v f = 40 ft/s; t = 10 s; a = 4 ft/s 2 ; ave v = 20 ft/s UNKNOWN: d = ? EQUATION : Ave v = d/t SOLVE : Ave v = d/t => d = (ave v)t SUBSTITUTION: d = (ave v)t = (20 ft/s)10 s = 200 ft

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Sample Problem #1 A brick falls freely from a high scaffold at a construction site. 1. What is the velocity after 4 seconds? 2. How far does the brick fall in this time?

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Solution Given: a = 9.8 m/s 2 t = 4s V f = 0 m/s + (-9.8 m/s 2 ) ( 4.0 s) = -39.2 m/s d = 0 m/s (4s) +.5(-9.8 m/s 2 ) (4s) 2 = 0 +.5(-9.8m/s 2 ) (16 s 2 ) = -78.4m What is the velocity after 4 seconds? Find: V How far does the brick fall in this time? Find: d

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Sample problem #2 An airplane must reach a speed of 71m/s for takeoff. If the runway is 1000m long, what must be the acceleration?

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Solution (71m/s) 2 = (0 m/s) 2 + 2 (a) ( 1000m) (-2000m) a = - 5041 m 2 / s 2 What is the acceleration needed to take off? Given: V i =0 m/sFind: a =? V f =71 m/s d=1000m a = 2.5 m/s 2

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Summary Determine the type of motion List the given information Choose the best formula from the Physics formulas Substitute the proper units Solve for the unknown in the equation

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GRAPHICAL REPRESENTATION OF VELOCITY slope - the slope of a displacement vs. time curve would be the velocity GBS Physics - position vs. time

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Constant Velocity Positive Velocity Changing Velocity (acceleration)

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Slow, Rightward (+) Constant Velocity Fast, Rightward (+) Constant Velocity http://www.physicsclassroom.com/mmedia/kinema/cpv.html

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Slow, Leftward (-) Constant Velocity Fast, Leftward (-) Constant Velocity http://www.physicsclassroom.com/mmedia/kinema/cnv.html

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GRAPHICAL REPRESENTATION OF ACCELERATION slope - the slope of a velocity vs. time curve would be the acceleration GBS Physics - velocity vs time http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/kinema/avd.html

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Positive Velocity Positive Acceleration Positive Velocity Zero Acceleration

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As previously learned, a plot of velocity-time can be used to determine the acceleration of an object (the slope). We will now learn how a plot of velocity versus time can also be used to determine the displacement of an object. For velocity versus time graphs, the area bound by the line and the axes represents the displacement. Determining the Area on a v-t Graph http://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L4e.html

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The shaded area is representative of the displacement during from 0 seconds to 6 seconds. This area takes on the shape of a rectangle can be calculated using the appropriate equation. Area = b * h Area = (6 s) * (30 m/s) Area = 180 m

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The shaded area is representative of the displacement during from 0 seconds to 4 seconds. This area takes on the shape of a triangle can be calculated using the appropriate equation. Area = 0.5 * b * h Area = (0.5) * (4 s) * (40 m/s) Area = 80 m

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The shaded area is representative of the displacement during from 2 seconds to 5 seconds. This area takes on the shape of a trapezoid can be calculated using the appropriate equation. Area = 0.5 * b * (h 1 + h 2 ) Area = (0.5) * (3 s) * (20 m/s + 50 m/s) Area = 105 m

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