Chapter 2. Motion in One Dimension

Transcription

1 Chapter 2 Motion in One Dimension

2 Motion in One Dimension Sections 2-1 Displacement and Velocity 2-2 Acceleration 2-3 Falling objects

3 Motion Displacement and Velocity One-dimensional motion is the simplest form Motion takes place over time and depends upon the frame of reference Frame of reference a coordinate system for specifying the precise location of objects in space Choose a reference frame and use it consistently Some reference frames simplify the problem

4 Displacement and Velocity Displacement Displacement is a change in position It is the straight-line distance (with direction) from the starting point to the ending point Displacement x = x f x i displacement = change in position = final position initial position SI Unit of Displacement: meter (m) Greek letter delta ( ) denotes a change in something

5 Displacement and Velocity Displacement Not always equal to the distance traveled Can be positive or negative (direction) positive directions: to the right; upward negative directions: to the left; downward

6 Displacement and Velocity Velocity Average velocity is displacement divided by the time interval Average Velocity v avg = x t = x f x i t f t i change in position average velocity = change in time SI Unit of Velocity: meter per second (m/s) = displacement time interval Positive or negative, depending on displacement

7 Displacement and Velocity Workbook Problem 2A The fastest fish, the sailfish, can swim km/h. Suppose you have a friend who lives on an island 16 km away from the shore. If you send a message using a sailfish as a messenger, how long will it take for the message to reach your friend?

8 Displacement and Velocity Velocity Velocity is not the same as speed Average speed tells how fast Average velocity tells how fast and what direction Velocity can be interpreted graphically If position vs. time is plotted on a graph, the average velocity can be found by finding the slope of the line

9 Velocity Displacement and Velocity Suppose a bicyclist is riding with constant velocity of v = +4 m/s. After riding for one second, his displacement is +4 m. Two seconds later, it is +12 m. The straight line in the graph shows his displacement x for any time t. Image credit: Cutnell & Johnson, Wiley Publishing, Physics 5 th Ed. Recall that v avg = x/ t. Notice that x/ t is the slope of the line in the position vs. time graph. So, the slope of the position vs. time graph is equal to the velocity.

10 Displacement and Velocity Image credit: Cutnell & Johnson, Wiley Publishing, Physics 5 th Ed. The graphs of the three segments of the bicyclist's trip each have different slopes. Where the slope is positive, the velocity is also positive. Zero slope means zero velocity. Negative slope indicates negative velocity.

11 Velocity Displacement and Velocity The graph of an accelerating object would look something like this one. The slope of the line tangent to a point is the instantaneous velocity (the velocity at that point in time). Image credit: Cutnell & Johnson, Wiley Publishing, Physics 5 th Ed.

12 Changes in Velocity Acceleration Acceleration measures the rate of change in velocity, or how quickly velocity changes Average Acceleration a avg = v t = v f v i t f t i change in velocity average acceleration = time required for change SI Unit of Acceleration: meter per second per second (m/s 2 ) Acceleration can be positive or negative and will agree with the direction of the change in velocity

13 Workbook Problem 2B Acceleration In 1977 off the coast of Australia, the fastest speed by a vessel on the water was achieved. If this vessel were to undergo an average acceleration of 1.80 m/s 2, it would go from rest to its top speed in 85.6 s. What was the speed of the vessel?

14 Changes in Velocity Acceleration Acceleration has direction and magnitude When v is negative, the acceleration is negative When v is positive, the acceleration is positive The slope and shape of the graph describe the object's motion The slope of the velocity vs. time graph yields average acceleration. Image credit: Cutnell & Johnson, Wiley Publishing, Physics 5 th Ed.

15 Acceleration Motion with Constant Acceleration Displacement depends on acceleration, initial velocity, and time ball moves top to bottom with constant acceleration (what is the cause?) velocity increases and so does distance velocity increases the same amount in each interval displacement increases the same amount in each interval

16 v avg = x t Displacement with Constant Uniform Acceleration and v avg = v i + v f 2 so x t = v avg = v i + v f 2 solving for x: x = 1 2 (v i + v f ) t displacement = 1 (initial velocity + final velocity)(time interval) 2

17 Workbook Problem 2C Acceleration In England, two men built a tiny motorcycle with a wheel base (the distance between the centers of the two wheels) of just 108 mm and wheel's measuring 19 mm in diameter. The motorcycle was ridden over a distance of 1.00 m. Suppose the motorcycle has constant acceleration as it travels this distance, so that its final speed is m/s. How long does it take the motorcycle to travel the distance of 1.00 m? Assume the motorcycle is initially at rest.

18 Acceleration Velocity with Constant Uniform Acceleration The average acceleration equation can be rearranged to yield a useful equation... ā = v t = v f v i t Velocity with Constant Uniform Acceleration v f = v i +ā t final velocity = initial velocity + (acceleration time interval) useful when displacement is not known

19 Acceleration Velocity with Constant Uniform Acceleration Substitute this expression for v f into the previous displacement equation... x = 1 2 (v i + v f ) t x = 1 2 (v i +(v i +ā t)) t Velocity with Constant Uniform Acceleration x = v i t a t2 displacement = (initial velocity time interval) + acceleration (time interval)2 1 2 useful when final velocity is not known

20 Workbook Problem 2D Acceleration

21 Workbook Problem 2D Acceleration

22 Workbook Problem 2D Acceleration v i = 15.0 m/s

23 Workbook Problem 2D Acceleration v i = 15.0 m/s v f = m/s

24 Workbook Problem 2D Acceleration v i = 15.0 m/s v f = m/s ā =+2.5 m/s 2

25 Workbook Problem 2D Acceleration v i = 15.0 m/s v f = m/s ā =+2.5 m/s 2 t =?

26 Workbook Problem 2D v i = 15.0 m/s v f = m/s ā =+2.5 m/s 2 t =? Acceleration

27 Acceleration Workbook Problem 2D v i = 15.0 m/s v f = m/s ā =+2.5 m/s 2 t =? ā = v f v i t appropriate equation

28 Acceleration Workbook Problem 2D v i = 15.0 m/s v f = m/s ā =+2.5 m/s 2 t =? ā = v f t = v f v i t v i ā = 15.0 ( 15.0) 2.5 = 12 s appropriate equation isolate unknown, sub & calc

29 Free Fall Falling Objects Freely falling bodies undergo constant acceleration All objects dropped near the surface of a planet fall with the same constant acceleration this is free fall (neglecting air resistance) With air resistance, objects do not fall at the same rate. Free-fall acceleration is denoted as g Near the surface of Earth, g = 9.81 m/s 2 Down is negative: a = g = 9.81 m/s 2

30 Falling Objects Free Fall What goes up must come down Gravity begins working on a thrown object as soon as it is released. Freely falling objects always have the same downward acceleration The force of gravity causes a downward acceleration on a freely falling object of 9.81 m/s 2

31 Workbook Problem 2F Falling Objects The famous Gateway to the West Arch in St. Louis,Missouri, is about 192 m tall at its highest point. Suppose Sally, a stuntwoman, jumps off the top of the arch. If it takes Sally 6.4 s to land on the safety pad at the base of the arch,what is her average acceleration? What is her final velocity?