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

Transcription

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

2 Defining Motion Motion is a continuous change in position can be described by measuring the rate of change of position with time. This chapter covers: Straight line Motion Uniform Circular Motion Projectile Motion

3 Speed Speed is a scalar quantity, equals the ratio of the total distance an object moves to the total time. the SI- unit of speed is meters per second (m/s). two ways to express the speed : average speed instantaneous speed.

4 Average speed Find the average speed? time = 4 min =2m =58m

5 Find the average speed?

6 Instantaneous Speed is the rate at which an object is moving at a given moment in time the speedometer in a car measures the car s instantaneous speed.

7 Vectors A Vector is a quantity described by a magnitude and direction Direction Origin, or application point Examples of vectors that will be used: Displacement, Velocity, Acceleration, Force Slide 3-61

8 Displacement and Distance Distance: is the actual path traveled between the initial and final position. Displacement : is a vector quantity between the initial and final position with direction toward the final position Section 2.2

9 Velocity Velocity ( ) is vector quantity, equals the ratio of the total vector displacement an object moves to the total time. Avelocity V = + 60 mi/ hr. V = - 60 mi/ hr.

10 Velocity Example Find the average velocity? total displacement Avergae velocity total time d 80 m v 20 m/s t 4.0 s Section 2.2

11 Velocity Example How far would the car travel in 10 s? Equation: v = d/t Rearrange equation: v.t = d d = (20 m/s).(10 s) = 200 m Section 2.2

12 Acceleration Acceleration the change in velocity during a given time Average acceleration change in velocity time for change to occur v vf v0 a (v f = final & v 0 = original) t t Units of acceleration = (m/s)/s = m/s 2 Section 2.3

13 Acceleration Acceleration can be generated in three ways: 1. Velocity increase in magnitude (speed up: acceleration) 2. Velocity decrease in magnitude (slow down: deceleration) Section 2.3

14 Acceleration 3. Change direction of velocity vector (turn, or circular )

15 Acceleration Example A race car starting from rest accelerates uniformly along a straight track, reaching a speed of 25 m/s in 7.0 s. Find its acceleration. Solution: a a vf v 0 t 25 m/s s 3.57 m/s 2 Section 2.3

16 Useful Equation Remember that a vf v t Rearrange this equation: at = v f v o v f = v o + at (solved for final velocity). This equation is very useful in computing final velocity 0 Section 2.3

17 Acceleration Example If the car in the preceding example 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 )? Solution: a = 3.57 m/s 2 (from preceding example) Use equation: v f = v o + at v f = 0 + (3.57 m/s 2 )(10 s) = 35.7 m/s

18 Do Heavier objects fall faster? vacuum moon inside a vacuum where air resistance is neglected, every freely falling object on Earth accelerates at the same rate, regardless of mass Section 2.3

19 Gravitational Acceleration = 9.8 m/s 2 The velocity of a free falling object increases uniformly at 9.80 m/s each second. The gravitational acceleration is constant and equals g = 9.80 m/s 2. Section 2.3

20 The distance traveled by the falling object increases each second, however this increase is not uniform Distance is proportional to t 2 d What about the distance a dropped object will travel? 1 2 gt Velocity is proportional to t v f = gt 2 Section 2.3

21 Free Fall example A ball is dropped from a tall building. How far does the ball drop in 1.50 s? Solution: Given: g = 9.80 m/s 2, t = 1.5 s d m/s 2 gt 1.5 s m/s s 2 11 m 2 2 Section 2.3

22 Solving for Final Speed Confidence Exercise What is the speed of the ball in the previous example 1.50 s after it is dropped? Solution: Given: g = a = 9.80 m/s 2, t = 1.5 s v f = v o + at = 0 + (9.80 m/s 2 )(1.5 s) = 14.7 m/s Section 2.3

23 Up and Down Gravity slows the ball, then speeds it up Acceleration due to gravity remains constant The ball returns to its starting point with the same speed it had initially: v o = v f Section 2.3

24 Which ball hits the ground first? An object thrown horizontally will fall at the same rate as an object that is dropped Section 2.5

25 The velocity in the horizontal direction does not affect the velocity and acceleration in the vertical direction Section 2.5

26 Trajectory of a projectile

27 Projected at an Angle (not horizontal) Combined horizontal vertical horizontal &vert components component component Section 2.5

28 In throwing a football the horizontal velocity remains constant but the vertical velocity changes like that of an object thrown upward. Section 2.5

29 If air resistance is neglected, projectiles have symmetric paths and the maximum range is attained at 45 o.

30 Acceleration in Uniform Circular Motion Although an object in uniform circular motion has a constant speed, it is constantly changing directions and therefore its velocity is constantly changing directions Since there is a change in direction there is a change in 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

31 Centripetal ( center-seeking ) Acceleration 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

32 Centripetal Acceleration a c v r 2 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

33 Finding Centripetal Acceleration Example Determine the magnitude of the centripetal acceleration of a car going 12 m/s on a circular track with a radius of 50 m. Given: v = 12 m/s, r = 50 m Solve: a c 2 2 v 12 m/s 2.9 m/s r 50 m Section 2.4

34 Finding Centripetal Acceleration Confidence Exercise Compute the centripetal acceleration in m/s 2 of the Earth in its nearly circular orbit about the Sun. Given: 11 4 Solve: r m, v m/s 4 v m/s m /s ac r m m a c = m/s Section 2.4

35 Important Equations Chapter 2 v d t (average speed) 1 2 d at (distance traveled, starting from rest) d gt (distance traveled, dropped object) 2 g = 9.80 m/s 2 = 32 ft/s 2 (acceleration, gravity) v vf v0 a (constant acceleration) t t v f = v o + at (final velocity with constant a) 2 v ac (centripetal acceleration) r Review

36 Chapter 2 Homework: