Linear Kinematics Chapter 2 in the text 5/2/2019 Dr. Sasho MacKenzie - HK 376

KINEMATICS LINEAR ANGULAR Future Class Scalars Distance Speed Future Class Vectors Displacement Velocity Acceleration 5/2/2019 Dr. Sasho MacKenzie - HK 376

Scalars A measure that only considers magnitude Does not consider direction E.g., a length of 2.7 meters is a scalar measure 2.7 m The javelin is 2.7 meters but has no direction 5/2/2019 Dr. Sasho MacKenzie - HK 376

Vectors Describes both a magnitude and direction E.g., a displacement of 15 meters in the positive direction is a vector. Represented by arrows, in which the length represents magnitude and orientation represents direction. The javelin meters in the positive direction 5/2/2019 Dr. Sasho MacKenzie - HK 376

Distance A measure of the length of the path followed by an object from its initial to final position. A scalar quantity (no direction) 5/2/2019 Dr. Sasho MacKenzie - HK 376

Speed The rate of motion of an object The rate at which an object’s position is changing. A scalar quantity (no direction) 5/2/2019 Dr. Sasho MacKenzie - HK 376

Displacement The straight-line distance in a specific direction from the starting position to the ending position. A vector quantity (must have direction) As the crow flies 5/2/2019 Dr. Sasho MacKenzie - HK 376

Velocity The rate of motion in a specific direction Similar to speed but with a direction A vector quantity 5/2/2019 Dr. Sasho MacKenzie - HK 376

Distance vs. Displacement W E S Distance vs. Displacement End 10 km Start 5 km 11.2 km 63.4° Distance 5 km + 10 km = 15 km Displacement Magnitude:  (5 km)2 + (10 km)2 = 11.2 km Direction: Inverse Tan (10 km / 5 km) = 63.4° East of North 5/2/2019 Dr. Sasho MacKenzie - HK 376

Speed vs. Velocity Speed Velocity It took Bill 3.5 hours in total to walk 5 km North and 10 km East. What was Bill’s average speed and average velocity? Speed = Distance/Time = 15 km/ 3.5 hours = 4.3 km/h Velocity = Displacement/Time = 11.2 km/ 3.5 hours = 3.2 km/h 63.4° E of N 5/2/2019 Dr. Sasho MacKenzie - HK 376

Acceleration The rate at which an object’s speed or velocity changes. When an object speeds up, slows down, starts, stops, or changes direction, it is accelerating. Always a vector quantity (has direction) 5/2/2019 Dr. Sasho MacKenzie - HK 376

Acceleration The direction of motion does not indicate the direction of acceleration. An object can be accelerating even if its speed remains unchanged. The acceleration could be due to a change in direction not magnitude. 5/2/2019 Dr. Sasho MacKenzie - HK 376

Circle Circumference = 2r; Circle Diameter = 2r; r is radius Sample Problem Bolt runs 200 m in 19.19 seconds. Assume he ran on the inside line of lane 1, which makes a semicircle (r = 36.5 m) for the first part of the race. He runs the curve in 11 s. Circle Circumference = 2r; Circle Diameter = 2r; r is radius What distance was run on the curve? What was his displacement after the curve? Total distance? Total displacement? Average velocity on the curve? Average speed on the curve? Average velocity for the race? Average speed for the race? Start N W E S 36.5 m Finish 5/2/2019 Dr. Sasho MacKenzie - HK 376

Instantaneous Velocity The average velocity over an infinitely small time period. Determined using calculus The derivative of displacement The slope of the displacement curve 5/2/2019 Dr. Sasho MacKenzie - HK 376

Instantaneous Acceleration The average acceleration over an infinitely small time period. Determined using Calculus The derivative of velocity The slope of the velocity curve 5/2/2019 Dr. Sasho MacKenzie - HK 376

Slope Y (4,8) 8 (0,0) X 4 Slope = rise = Y = Y2 – Y1 = 8 – 0 = 8 = 2 Slope could be anything not just velocity: give other example 4 Slope = rise = Y = Y2 – Y1 = 8 – 0 = 8 = 2 run X X2 – X1 4 – 0 4 5/2/2019 Dr. Sasho MacKenzie - HK 376

Velocity is the slope of Displacement (4,8) 8 Displacement (m) (0,0) X 4 Time (s) Average Velocity = rise = D = D2 – D1 = 8 – 0 = 8 m = 2 m/s run t t2 – t1 4 – 0 4 s 5/2/2019 Dr. Sasho MacKenzie - HK 376

The displacement graph on the previous slide was a straight line, therefore it’s slope was 2 at every instant. Which means the velocity at any instant is equal to the average velocity. However if the graph was not straight the instantaneous velocity could not be determined from the average velocity. 5/2/2019 Dr. Sasho MacKenzie - HK 376

Average vs. Instantaneous X Y 8 4 Displacement (m) Time (s) The average velocity does not accurately represent slope at this particular point. (0,0) (4,8) Average Velocity = rise = D = D2 – D1 = 8 – 0 = 8 m = 2 m/s run t t2 – t1 4 – 0 4 s Read Ch. 6 pages 147-158 for next class 5/2/2019 Dr. Sasho MacKenzie - HK 376

Circle Circumference = 2r; Circle Diameter = 2r; r is radius Sample Problem Bolt runs 200 m in 19.19 seconds. Assume he ran on the inside line of lane 1, which makes a semicircle (r = 36.5 m) for the first part of the race. He runs the curve in 11 s. Circle Circumference = 2r; Circle Diameter = 2r; r is radius What distance was run on the curve? What was his displacement after the curve? Total distance? Total displacement? Average velocity on the curve? Average speed on the curve? Average velocity for the race? Average speed for the race? 114.7 m 73 m South Start 200 m 112.3 m 40.5° South of East N W E S 6.64 m/s South 36.5 m 10.424 m/s 5.85 m/s 40.5° South of East 10.422 m/s Finish 5/2/2019 Dr. Sasho MacKenzie - HK 376