Foundations of Physics CPO Science Foundations of Physics Chapter 9 Unit 1, Chapter 4

Unit 1: Measurement and Motion Chapter 4: Acceleration in a Straight Line 4.1 Acceleration 4.2 A Model for Accelerated Motion 4.3 Free Fall and the Acceleration due to Gravity

Chapter 4 Objectives Calculate acceleration from the change is speed and the change in time. Give an example of motion with constant acceleration. Determine acceleration from the slope of the speed versus time graph. Calculate time, distance, acceleration or speed when given three of the four variables. Solve two-step accelerated motion problems. Calculate height, speed, or time of flight in free fall problems. Explain how air resistance makes objects of different masses fall with different accelerations.

Chapter 4 Vocabulary Terms initial speed free fall acceleration due to gravity (g) time of flight friction air resistance terminal speed acceleration m/sec2 delta D constant acceleration uniform acceleration slope term

4.1 Acceleration Key Question: How is the speed of the ball changing? *Students read Section 4.1 AFTER Investigation 4.1

4.1 Acceleration of a car Acceleration is the rate of change in the speed of an object.

4.1 Acceleration vs. Speed Positive acceleration and positive speed

4.1 Acceleration vs. Speed Negative acceleration and positive speed

Change in speed (m/sec) 4.1 Acceleration Change in speed (m/sec) Acceleration (m/sec2) a = Dv Dt Change in time (sec)

4.1 Calculate Acceleration A student conducts an acceleration experiment by coasting a bicycle down a steep hill. The student records the speed of the bicycle every second for five seconds. Calculate the acceleration of the bicycle. 1) You are asked for the acceleration. 2) You are given a table of speeds and times from an experiment. 3) a = (v2-v1) ÷ (t2 - t1) 4) Choose any two pairs of speed and time data. a = (6 m/sec - 4 m/sec) ÷ (3 sec - 4 sec) = 2 m/sec2 For this experiment, the acceleration would have been the same for any two points.

4.1 Acceleration and Speed Constant acceleration is different from constant speed. Motion with zero acceleration appears as a straight horizontal line on a speed versus time graph. zero acceleration constant speed

4.1 Acceleration and Speed Constant acceleration is sometimes called uniform acceleration. A ball rolling down a straight ramp has constant acceleration. constant acceleration increasing speed

4.1 Acceleration and Speed An object can have acceleration, but no speed. Consider a ball rolling up a ramp. As the ball slows down, eventually its speed becomes zero. constant negative acceleration decreasing speed

4.1 Slope and Acceleration Use slope to recognize when there is acceleration in speed vs. time graphs. Level sections (A) on the graph show an acceleration of zero. The highest acceleration (B) is the steepest slope on the graph. Sections that slope down (C) show negative acceleration (slowing down).

4.2 A Model for Accelerated Motion Key Question: How do we describe and predict accelerated motion? *Students read Section 4.2 AFTER Investigation 4.2

4.2 Slope of a graph The slope of a graph is equal to the ratio of rise to run. On the speed versus time graph, the rise and run have special meanings, as they did for the distance versus time graph. The rise is the amount the speed changes. The run is the amount the time changes.

4.2 Acceleration and slope Acceleration is the change in speed over the change in time. The slope of the speed versus time graph is the acceleration.

4.2 Calculate acceleration The following graph shows the speed of a bicyclist going over a hill. Calculate the maximum acceleration of the cyclist and say when in the trip it occurred. 1) You are asked for the acceleration. 2) You are given a graph of speed versus time. 3) a = slope of graph 4) The steepest slope is between 60 and 70 seconds, when the speed goes from 2 to 9 m/sec. a = (9 m/sec - 2 m/sec) ÷ (10 sec) = 0.7 m/sec2

4.2 Solving Motion Problems

4.2 Solving Motion Problems

4.2 Calculate speed A ball rolls at 2 m/sec onto a ramp. The angle of the ramp creates an acceleration of 0.75 m/sec2. Calculate the speed of the ball 10 seconds after it reaches the ramp. 1) You are asked for the speed. 2) You are given an initial speed acceleration, and time. 3) v = vo + at 4) v = 2 m/sec + (.75 m/sec2)(10 sec) = 9.5 m/sec

4.2 Solving Motion Problems initial position distance if at constant speed distance to add or subtract, depending on acceleration

4.2 Calculate position A ball traveling at 2 m/sec rolls onto a ramp that tilts upward. The angle of the ramp creates an acceleration of -0.5 m/sec2. How far up the ramp does the ball get at its highest point? (HINT: The ball keeps rolling upward until its speed is zero.) 1) You are asked for distance. 2) You are given an initial speed and acceleration. You may assume an initial position of 0. 3) v = vo + at x = xo + vot + 1/2at2 4) At the highest point the speed of the ball must be zero. 0 = 2 m/sec - 0.5t t = 4 seconds Now use the time to calculate how far the ball went. x = (2 m/sec)(4 sec) - (0.5) (.5 m/sec2) (4 sec)2 = 4 meters At its highest point, the ball has moved 4 meters up the ramp.

4.2 Solving Motion Problems

4.2 Calculate time A car at rest accelerates at 6 m/sec2. 1) You are asked for time and speed. 2) You are given vo = 0, x = 440m, and a = 6 m/sec2 assume xo = 0 3) v = vo + at x = xo + vot + 1/2at2 4) Since xo = vo = 0, the position equation reduces to: x = 1/2at2 440 m = - (0.5)(6 m/sec2) t2 t2 = 440 ÷ 3 = 146.7 t = 12.1 seconds Now use the time to calculate the speed. v = (6 m/sec2)(12.1 sec) = 72.6 m/sec This is 162 miles per hour. A car at rest accelerates at 6 m/sec2. How long does it take to travel 440 meters (about a quarter-mile) and how fast is the car going at the end?

4.2 Calculate position A ball starts to roll down a ramp with zero initial speed. After one second, the speed of the ball is 2 m/sec. How long does the ramp need to be so that the ball can roll for 3 seconds before reaching the end? 1) You are asked to find position (length of the ramp). 2) You are given vo = 0, v = 2m/sec at t = 1 sec, t = 3 sec at the bottom of the ramp, and you may assume xo = 0. 3) After canceling terms with zeros, v = at and x = 1/2at2 4) This is a two-step problem. First, you need the acceleration, then you can use the position formula to find the length of the ramp. a = v ÷ t = (2 m/sec) ÷ (1 sec) = 2 m/sec2 x = 1/2at2 = (0.5)(2 m/sec2)(3 sec)2 = 9 meters

4.3 Solving Problems with Free Fall

4.3 Calculate height A stone is dropped down a well and it takes 1.6 seconds to reach the bottom. How deep is the well? You may assume the initial speed of the stone is zero. 1) You are asked for distance. 2) You are given an initial speed and time of flight. 3) v = vo - gt y = yo + vot - 1/2gt2 4) Since yo = vo = 0, the height equation reduces to: y = - 1/2gt2 y = - (0.5)(9.8 m/sec2) × (1.6 sec)2 y = -12.5 meters The negative sign indicates the height is lower than the initial height by 12.5 m.

4.3 Air Resistance and Mass The acceleration due to gravity does not depend on the mass of the object which is falling. Air creates friction that resists the motion of objects moving through it. All of the formulas and examples discussed in this section are exact only in a vacuum (no air).

4.3 Terminal Speed You may safely assume that a = g = 9.8 m/sec2 for speeds up to several meters per second. The resistance from air friction increases as a falling object’s speed increases. Eventually, the rate of acceleration is reduced to zero and the object falls with constant speed. The maximum speed at which an object falls when limited by air friction is called the terminal speed.

4.3 Free Fall and Acceleration due to Gravity Key Question: How do you measure the acceleration of a falling object? *Students read Section 4.3 BEFORE Investigation 4.3

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