Would you risk your life driving drunk? Intro

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1 Martha Casquete

3 Would you risk your life driving drunk? Intro

4 Assignments: For next class: Finish reading Ch. 2, read Chapter 3 (Vectors) HW3 Set due next Wednesday, 9/11 HW3 will be in weebly. Question/Observation Mondays Research Q/O Wednesday with HW (due date Wednesday)

5 Motion Speed and Velocity Acceleration Acceleration in Uniform Circular Motion Projectile Motion What today s lesson has to do with drunk drivers?

6 Units: Length, Mass, and Time Order of magnitude calculations Dimensional Analysis Conversion of Units Significant digits (on your own)

7 Motion is everywhere walking, driving, flying, etc. This chapter focuses on definition/discussion of: speed, velocity, and acceleration. There are two basic kinds of motion: Straight line Circular Section 2.1

8 Position the location of an object A reference point must be given in order to define the position of an object Motion an object is undergoing a continuous change in position Description of Motion the time rate of change of position A combination of length and time describes motion Section 2.1

9 Any motion involves three concepts Displacement/distance Velocity/speed Acceleration

10 Displacement is a change of position in time. It is a vector quantity. It has both magnitude and direction: + or - sign It has units of [length]: meters. x 1 (t 1 ) = m x 2 (t 2 ) = m Δx = -2.0 m m = -4.5 m x 1 (t 1 ) = m x 2 (t 2 ) = m Δx = +1.0 m m = +4.0 m

11 Sometimes displacement is confused with distance. It is important that you understand the difference between distance and displacement. Distance the actual path length traveled If I start walking when x 1 = - 3.0m and then I return to the same place, then x 2 = - 3.0m. a) What is my displacement? b) What is my distance? a) What is my displacement? x 1 = m x 2 = m Δx = -3.0 m (- 3.0 m) = 0 m b) What is my distance? x 1 = 3.0 m x 2 = 3.0 m x 1 + x 2 = 3.0 m m = 6.0m

12 Every day, I travel a distance of 108 miles from Brownsville/Rancho Viejo to Edinburg and back. Every day, at 8:20pm I noticed that I have displaced myself zero miles. Where am I? What is the distance between one side of the wall and the other in the classroom? If I pace it, what will be my displacement? Displacement is not Distance.

13 In Physical Science speed and velocity have different (distinct) meanings. Speed a scalar quantity, only magnitude (numerical value and unit measurement)/scalar A car going 80 km/h Velocity vector, both magnitude & direction A car going 80 km/h north Section 2.2

14 Vector quantities may be represented by arrows. The length of the arrow is proportional to magnitude. 40 km/h 80 km/h -40 km/h -80 km/h Section 2.2

15 Note that vectors may be both positive and negative. Section 2.2

16 distance traveled Average Speed = time to travel distance v = d/t or v = Dd/Dt ( where D means change in ) Over the entire time interval, speed is an average Instantaneous Speed the speed of an object at an instant of time (Dt is very small) Glance at a speedometer Section 2.2

17 Section 2.2

18 Velocity is similar to speed except a direction is involved. Average velocity = displacement total travel time Instantaneous velocity similar to instantaneous speed except it has direction Section 2.2

19 What is the speed of a car if it has traveled 100m in 4.0 s. GIVEN: d = 1000 m & t = 4.0 s EQUATION: v = d/t SOLVE: d/t = 100 m/4.0 s = 25 m/s = average speed Velocity would be described as 25 m/s in the direction of motion (east?) Section 2.2

20 How far would the car above travel in 10s? EQUATION: v = d/t REARRANGE EQUATION: vt = d SOLVE: (20 m/s) (10 s) = 200 m Section 2.2

21 (it is on your book do the math) Section 2.2

22 GIVEN: t = 365 days (must convert to hours) Earth s radius (r) = 9.30 x 10 7 miles (p. 19) CONVERSION: t = 365 days x (24h/day) = 8760 h MUST FIGURE DISTANCE (d): d =? Recall that a circle s circumference = 2pr d=2pr (and we know p and r!!) Therefore d = 2pr = 2 (3.14) (9.30 x 10 7 mi) Section 2.2

SOLVE EQUATION: v = d/t d/t = 2 (3.14) (9.30 x 10 7 mi) = 0.

23 SOLVE EQUATION: v = d/t d/t = 2 (3.14) (9.30 x 10 7 mi) = x 10 7 mi/h 8760 h ADJUST DECIMAL POINT: v = avg. velocity = 6.67 x 10 4 mi/h = 66,700 mi/h Section 2.2

24 Radius of the Earth = R E = 4000 mi Time = 24 h Circumference of Earth = 2pr = 2 (3.14) (4000mi) Circumference of Earth = 25,120 mi v = d/t = (25,120 mi)/24h = 1047 mi/h Section 2.2

25 Changes in velocity occur in three ways: Increase in magnitude (speed up) Decrease in magnitude (slow down) Change direction of velocity vector (turn) When any of these changes occur, the object is accelerating. Faster the change Greater the acceleration Acceleration the time rate of change of velocity Section 2.3

26 A measure of the change in velocity during a given time period Avg. acceleration = Dv a = = t v f v o t change in velocity time for change to occur (v f = final & v o = original) Units of acceleration = (m/s)/s = m/s 2 In this course we will limit ourselves to situations with constant acceleration. Section 2.3

27 As the velocity increases, the distance traveled by the falling object increases each second. Section 2.3

28 v f v Remember that a = o t Rearrange this equation: at = v f v o v f = v o + at velocity) (solved for final This equation is very useful in computing final velocity Section 2.3

29 If a car with an acceleration of 3.57 m/s 2 continues to accelerate at the same rate for three more seconds, what will be the magnitude of its velocity in m/s at the end of this time (v f )? a = 3.57 m/s 2 (found in preceding example) t = 10 s v o = 0 (started from rest) Use equation: v f = v o + at v f = 0 + (3.57 m/s 2 ) (10 s) = 35.7 m/s Section 2.3

30 Acceleration (+) Deceleration (-) Section 2.3

31 Special case associated with falling objects Vector towards the center of the earth Denoted by g g = 9.80 m/s 2 Section 2.3

Astronaut David Scott demonstrated the principle on the moon, simultaneously dropping a feather and a hammer.

32 If frictional effects (air resistance) are neglected, every freely falling object on earth accelerates at the same rate, regardless of mass. Galileo is credited with this idea/experiment. Astronaut David Scott demonstrated the principle on the moon, simultaneously dropping a feather and a hammer. Each fell at the same acceleration, due to no atmosphere & no air resistance Copyright Houghton Mifflin Company. All rights reserved. Section 2.3

33 Galileo is also credited with using the Leaning Tower of Pisa as an experiment site. Section 2.3

34 d = ½ gt 2 This equation will compute the distance (d) an object drops due to gravity (neglecting air resistance) in a given time (t) Section 2.3

35 A ball is dropped from the top of an 80 m building. Does it reach the ground in 4.0s? GIVEN: g = 9.80 m/s 2, t = 4.0 s SOLVE: d = ½ gt 2 = ½ (9.80 m/s 2 ) (4.0 s) 2 = ½ (9.80 m/s 2 ) (16.00 s 2 ) =?? m d = 78 m Section 2.3

36 What is the speed of the ball in the previous example 4 s after it is dropped? Use equation: v f = v o + at v o = 0 GIVEN: g = a = 9.80 m/s 2, t = 4 s SOLVE: v f = at = (9.80 m/s 2 )(4 s) Speed of ball after 4 seconds = 39.2 m/s Section 2.3

37 Distance is proportional to t 2 (d = ½ gt 2 ) Velocity is proportional to t (v f = at ) Section 2.3

38 Acceleration due to gravity occurs in BOTH directions. Going up (-) Coming down (+) The ball returns to its starting point with the same speed it had initially. v o = v f Section 2.3

39 Although an object in uniform circular motion has a constant speed, it is constantly changing directions and therefore its velocity is constantly changing. Since there is a change in direction there is an acceleration. What is the direction of this acceleration? It is at right angles to the velocity, and generally points toward the center of the circle. Section 2.4

40 Supplied by friction of the tires of a car The car remains in a circular path as long as there is enough centripetal acceleration. Section 2.4

41 a c = v 2 r This equation holds true for an object moving in a circle with radius (r) and constant speed (v). From the equation we see that centripetal acceleration increases as the square of the speed. We also can see that as the radius decreases, the centripetal acceleration increases. Section 2.4

42 Determine the magnitude of the centripetal acceleration of a car going 12 m/s on a circular cloverleaf with a radius of 50 m. Section 2.4

43 Determine the magnitude of the centripetal acceleration of a car going 12 m/s on a circular cloverleaf with a radius of 50 m. GIVEN: v = 12 m/s, r = 50 m SOLVE: v a c = 2 (12 m/s) = 2 = 2.9 m/s r 50m 2 Section 2.4

44 Compute the centripetal acceleration in m/s 2 of the Earth in its nearly circular orbit about the Sun. GIVEN: r = 1.5 x m, v = 3.0 x 10 4 m/s SOLVE: v a c = 2 (3.0 x 10 = 4 m/s) 2 = r 1.5 x m a c = 6 x 10-3 m/s x 10 8 m 2 /s x m Section 2.4

45 An object thrown horizontally will fall at the same rate as an object that is dropped. Multiflash photograph of two balls Section 2.5

46 2-46 Section

47 An object thrown horizontally combines both straight-line and vertical motion each of which act independently. Neglecting air resistance, a horizontally projected object travels in a horizontal direction with a constant velocity while falling vertically due to gravity Section 2.5

48 Combined Horz/Vert. Components = Vertical Component + Horizontal Component Section 2.5

49 2-49 Section 2.5

51 2-51 Section 2.5

Angle of release Spin on the ball Size/shape of ball/projectile Speed of wind Direction of wind Weather conditions (humidity, rain, snow, etc) Field altitude (how much

52 Angle of release Spin on the ball Size/shape of ball/projectile Speed of wind Direction of wind Weather conditions (humidity, rain, snow, etc) Field altitude (how much air is present) Initial horizontal velocity (in order to make it out of the park before gravity brings it down, it must leave the bat at a high velocity) 2-52 Section 2.5

53 v = d/t (average speed) d = ½at 2 (distance traveled, starting from rest) d = ½gt 2 (distance traveled, dropped object) g = 9.80 m/s 2 = 32 ft/s 2 (acceleration, gravity) Dv t v f v o t a = = (constant acceleration) v f = v o + at (final velocity with constant a) a c = v 2 /r (centripetal acceleration) 2-53 Review

James T. Shipman Jerry D. Wilson Charles A. Higgins, Jr. Omar Torres. Chapter 2 Motion Cengage Learning

James T. Shipman Jerry D. Wilson Charles A. Higgins, Jr. Omar Torres Chapter 2 Motion Defining Motion Motion is a continuous change in position can be described by measuring the rate of change of position