1. Motion: Kinematics Chapter 2 Mechanics - the study of the motion of objects.  Kinematics is the science of describing the motion of objects using words, diagrams, numbers, graphs, and equations. Kinematics is a branch of mechanics.

2. Physics is a mathematical science. The underlying concepts and principles have a mathematical basis. Throughout the course of our study of physics, we will encounter a variety of concepts which have a mathematical basis associated with them.

3. I. The Nature of Physical Quantities: Vectors Vs. Scalars A. Scalars A scalar quantity is one that can be described by a magnitude (NUMERICAL value) alone. Ex: temperature, time, mass, volume

4. B. Vectors Vector quantities have both a MAGNITUDE (NUMERICAL value) and a DIRECTION. Arrows are used to represent vector quantities. The direction of the arrow gives the direction of the vector, and the length of the arrow is proportional to the magnitude of the vector. In other words, the longer the arrow, the greater the magnitude of the vector.

6. II. Motion is relative to the point it is taken from Car driving at 55 mph Earth revolves around the Sun at 66,000 mph Galaxy spins around its center at millions of mph Assume observations are from an observer on Earth unless otherwise stated A FRAME OF REFERENCE is an object that is used for comparison against.

7. Ex: Bob is standing by the pencil sharpener. The frame of reference is the pencil sharpener since that is what you are comparing Bob’s position against. Compare your position to those of other objects.

8. III. Total Distance Vs. Displacement A. Total Distance – all movement added up. B. Displacement (x)- straight line segment that connects the initial and final positions. Displacement is also a vector quantity and therefore needs a magnitude and direction. Δx = Change in Displacement = x f - x i x f = final displacement x i = initial displacement It can be a negative value- which means a movement to the left.

9. Ex: If I run around a 400 m track, what is my total distance covered? Distance = 400 m Displacement = 0m Since x f and x i are the same point Ex: If I ran from the 20 yd line to the 40 in football, what was my displacement? Δx = x f - x i = 40 - 20 = 20 yds If the other team ran from the 40 yd line to the 10, what was their displacement? Δx = x f - x i = 10 - 40 = -30 yds

10. IV. Types of Motion A. Average Speed = Total Distance ÷ Elapsed Time Rate at which distance is covered Measured in distance per unit time SI Unit: m/s Ex: miles/hour; meters/second When doing motion problems, place all final answer in their SI forms unless told otherwise.

11. Problem: While driving to Kona to buy Buta some kitty litter, Mrs. Takata was able to maintain a speed of 60.0 mph for 15.0 minutes. How many feet did she travel? 1mi = 5,280 ft s = d / t d = s (t) = 60 miles x 0.25 hr hr Speed = 60.0 mph Time = 15.0 minutes 15 miles x 5280 ft = mi 79,200 ft

12. Ex: How far does a jogger run in 1.5 hours if his average speed is 2.22 m/s? ( ) t = 1.5 h 60 min 60 s 1 hr 1 min = 5400 s s = 2.22 m/s s = d/t So, d = s (t) d = 2.22 m/s (5400 s) d = 12,000 m

13. Ex: With an average speed of 67 m/s, how long does it take a falcon to dive to the ground along a 150 m path? s = 67 m/s d = 150 m t = d ÷ s t = 150 m ÷ 67 m/s t = 2.2 s

14. B. Instantaneous Speed Speed at one given moment where time approaches zero. It cannot reach zero since you cannot divide by zero! (t lim  0) C. Velocity = Speed in a given DIRECTION (Contains a speed and a direction) Ex: 50 mph, North (this is a vector since it contains a magnitude and direction) v i = initial velocity v f = final velocity

15. D. Velocity = change in displacement; direction change in time = Δx; direction Δt SI Unit: m/s Average Velocity = change in velocity; direction 2 = Δv; direction 2 = v i + v f ; direction 2

16. E. Acceleration = Change in velocity over time = Δv; direction Δt SI Unit: m/s 2 m/s 2 = the change in velocity in m/s for each s that the object travels. In other words, how the velocity changes in m/s every second.

17. Negative Velocity and Negative Acceleration Negative Velocity: If you do a velocity calculation and end up with a negative answer, that means that the object is moving in the opposite direction. This is indicated by the direction of the vector. Ex: If we set “up” or “right” as the positive direction, we say that the object is going to the “down” or “left” when it has a negative velocity.

18. Velocity Time Velocity Time Velocity increases over time = positive acceleration Velocity decreases over time = negative acceleration Negative Acceleration: If you do an acceleration Calculation and end up with a negative answer, that means that the object is slowing down. In physics we say that the object is experiencing“negative acceleration” instead of saying that it is decelerating. However, recently, both terms are being used interchangeably.

19. Ex: A car travels 3.00 km due west, then 4.00 km due south. What is the car’s displacement? Start Finish θ haha hoho h h = √ (h a 2 + h o 2 ) h = √ ( (3.00 km) 2 + (4.00 km) 2 ) h = 5.00 km θ = tan -1 h o h a ( ) θ = tan -1 4.00 km 3.00 km ( ) h a = 3.00 km; west h o = 4.00 km; south θ = 53.1° South of West

20. In the same example, the car takes 10.0 min to travel from its initial to final position, what is its AVERAGE SPEED? Average speed = total distance ÷ elapsed time d = 3.00 km + 4.00 km = 7.00 km t = 10.0 min60 s 1 min ( ) = 600 s 7.00 km 1000 m 1 km ( ) = 7,000 m Average speed = 7,000 m ÷ 600 s = 11.7 m/s

21. In the same example, find the AVERAGE VELOCITY of the car. v = x t x = 5.00 km; 53.1° S of W v = 5,000 m 600 s x = 5.00 km1000 m 1 km ( ) = 5,000 m t = 600 s = 8.33 m/s; 53.1° S of W In general, the avg speed is not the same as the magnitude of the avg velocity. The one special case where these two are the same is when an object travels in a straight line and DOES NOT change the direction of its motion!

22. Ex: A car is traveling at 10 m/s in a straight line speeds up to 30 m/s in 5.0 s. What is the average acceleration of the car? v i = 10 m/s v f = 30 m/s t = 5.0 s a = Δv; direction Δt = v f – v i t = 30 m/s – 10 m/s 5.0 s = 4 m/s s = 4 m/s 2 For every second that the car sped up, its velocity increased by 4 m/s

23. Ex: A truck slows down from 25 mi/hr to 10 mi/hr in 0.03 hr. What was the trucks acceleration(or negative acceleration) in mi/hr 2 ? a = Δv; direction Δt = v f – v i t = 10 mi/hr – 25 mi/hr 0.03 hr = - 15 mi/hr 0.03 hr v i = 25 mi/hr v f = 10 mi/hr t = 0.03 hr = - 500 mi/hr 2 = 500 mi/hr 2 ; decelerating A negative acceleration means that the object is slowing down instead of speeding up.

24. V. Kinematic Equations: The following are true for motion where the acceleration is CONSTANT a = constant v f = v i + at x f = x i + v i t + ½ at 2 v f 2 = v i 2 + 2a(x f - x i ) NOTE: x i (starting position) and v i (starting velocity) often equal zero. The object starts from rest. Some books give more equations, but they can be simplified into these.

25. Ex: Reyn was driving in front of the school when he saw Mrs. Takata walk in front of him. Not wanting to hit Mrs. Takata, he came to a screeching halt by decelerating 12.5 m/s 2 in only 15 m. What was his velocity before he started breaking, in m/s, and how many seconds did it take him to stop? What we know: a = - 12.5 m/s 2 Δx = 15 m v f = 0 m/s What we don’t know: t = ? v i = ? Let’s look at our equations to see which one to use: v f = v i + at We don’t know v i and t x f = x i + v i t + ½ at 2 We don’t know v i and t v f 2 = v i 2 + 2a(x f - x i ) We know all of these!!!

26. 0 = v i 2 + 2(-12.5 m/s 2 ) (15 meters) -V i 2 = -375 m 2 /s 2 V i = 19.36 m/s = v f 2 = v i 2 + 2a(x f - x i ) v f = v i + at 0 = 19 m/s + -12.5 m/s 2 (t) -19 m/s = -12.5 m/s 2 (t) (t) = 1.52 sec 19 m/s Now, we solve for the time: t = 1.5 s

27. Ex: Mrs. Cama was racing to be the last customer in Ross before they closed for the night. If she is moving at 4.50 m/s and was only 10.0 m from the door, can she make it in before they close in 2.00 seconds? She can accelerate at 2.00 m/s 2 or can decelerate at 1.00 m/s 2. Should she try to rush in or stop? Hint: Where does she end up if she runs in? Where does she end up if she stops? If she runs in: v i = 4.50 m/s x i = 0 m (starting point) t = 2.00 s a = 2.00 m/s 2 If she stops: v i = 4.50 m/s x i = 0 m (starting point) t = 2.00 s a = - 1.00 m/s 2 (- since its decel)

28. If she runs in: x f = x i + v i t + ½ at 2 Since it asks, “Where does she end up if she runs in? “ x f = 0 + (4.5 m/s)(2 s) + ½ (2 m/s 2 )(2 s) 2 x f = 9 m + 4 m = 13 meters So if she runs, she runs 13 m Before the door closes! (RUN CAMA, RUN!)

29. If she stops: v f 2 = v i 2 + 2a(x f – x i ) Note: We don’t choose the equation x f = x i + v i t + ½ at 2 Because, we don’t want to know where she is after 2 seconds if she is slowing down, we want to know where she is when she stops! (notice she Doesn’t stop after 2 seconds, she’s still slowing down so for all we know she may still stop before she hits the door, or she may not???) 0 = ( 4.5 m/s) 2 + (2) (-1 m/s 2 )(Δx) v f 2 = v i 2 + 2a(x f – x i ) 0 = 20.25 m 2 /s 2 + -2m/s 2 (Δx) -20.25 m 2 /s 2 = -2m/s 2 (Δx) Δx = 10.125 m So even if she tries to stop, she will still hit the door!

30. Ex: A car initially traveling at 7.0 m/s accelerates uniformly at the rate of 0.80 m/s 2 for a distance of 245 m. a)What is its velocity at the end of the acceleration? b)What is its velocity after it accelerates for 125 m? What we know: v i = 7.0 m/s a = 0.80 m/s 2 Δx = 245 m a) Use: v f 2 = v i 2 + 2a(x f – x i )We don’t know the time v f 2 = (7 m/s) 2 + 2(0.8 m/s 2 )(245m) v f 2 = 49 m 2 /s 2 + 392 m 2 /s 2 v f 2 = 441 m 2 /s 2 v f = 21 m/s b) Substitute 125 for 245m v f = 16 m/s

31. VI. Falling objects – all objects fall (or decelerate if going up) with the same acceleration due to the fact that Gravity is making them fall. a = g = 9.80 m/s 2 We say that gravity naturally goes downward. So, we set g = + 9.80 m/s 2 in the down direction and g = - 9.80 m/s 2 in the upward direction. g = - 9.80 m/s 2 g = (+) 9.80 m/s 2

32. We use the same equations but usually we substitute y for x since we’re moving in a vertical motion instead of horizontal v f = v i + at In the horizontal direction In the vertical direction x f = x i + v i t + ½ at 2 v f 2 = v i 2 + 2a(x f - x i ) v f = v i + gt y f = y i + v i t + ½ gt 2 v f 2 = v i 2 + 2g(y f - y i )

33. Ex: Objects dropped: If you drop a rock off of R-building, 15 meters, how long does it take to hit the ground? How fast is it going when it hits the ground in m/s? y f = y i + v i t + ½ gt 2 15m = 0 + 0 + ½ (9.8m/s 2 )(t 2 ) 15m = 4.9m/s 2 (t 2 ) 3.06 s 2 = t 2 t = 1.7 seconds y f = 15 m y i = 0 m v i = 0 m/s g = 9.8m/s 2

34. Ex: Objects thrown down: If you threw a water balloon off the roof of your hotel 140 meters high with an initial velocity of 5.0 m/s, how fast would they be going, in m/s, when they hit the ground and how long would it take to hit the ground? v f 2 = v i 2 + 2g(Δy) v f 2 = (5m/s) 2 + 2(9.80m/s 2 )(140m) v f 2 = 25m 2 /s 2 + 2744m 2 /s 2 v f 2 = 2769 m 2 /s 2 v f = 53 m/s v f = v i + gt 53m/s = 5m/s + (9.80m/s 2 )(t) 48m/s = (9.80m/s 2 )(t) 4.9 s = t Δy = 140 m v i = 5.0 m/s g = 9.80m/s 2

35. Ex: Objects Thrown upwards: If I threw a baseball 25 m/s into the air, how high would it go, in meters, and how long would it be in the air? v f 2 = v i 2 + 2g(Δy) v i = 25 m/s g = - 9.80m/s 2 v f = 0 m/s (when an object is thrown upward, the velocity at the apex is 0 m/s because it lingers there for a split second before it starts to fall in the opposite direction) 0 = (25 m/s) 2 + 2(-9.8m/s 2 )( Δy) = -32 m Δy = 32 m; up

36. y f = y i + v i t + ½ gt 2 y f = 0 + 25m/s(t) + (-4.9m/s 2 )(t 2 ) y f = 25t – 4.9t 2 y f = t(25 - 4.9t) t = 0 & t = 5.1 sec