1. Accelerated Motion Merrill Physics Principles and Problems

2. Chapter Objectives  Relate velocity and acceleration to the motion of objects.  Draw and interpret velocity-time graphs.  Determine mathematical relationships among position, velocity, acceleration, and time.  Solve constant-acceleration problems using graphs and mathematical relationships.  Define free fall.  Solve free fall problems.

3. Velocity-Time Graphs  What does the slope of a velocity-time graph indicate?

4. Change in Velocity!!  Acceleration is any change in velocity  Example: What are the three controls in a vehicle that cause it to accelerate?

5.  A. Acceleration  1. the change in velocity divided by time  2. Average Acceleration = a =  v /  t delta stands for “change in”  3. units - meter per second squared - m/s 2 1 m / s 2 means velocity changes 1 m / s for every second that goes by  4. vector quantity - so associate a sign with direction

6.  5. Negative acceleration does not always indicate deceleration  a. just indicates that acceleration is in the negative direction.  b. you have deceleration when the signs of velocity and acceleration are opposites  6. Uniform Acceleration  a. When the acceleration remains constant over a period of time  b. found in nature in a number of circumstances - especially ‘g’

7.  7. Instantaneous Acceleration  a. Change in velocity at an instant in time  b. slope of a tangent to a velocity-time graph at than instant in time

8.  8. Example - The velocity of an automobile increases from 0 to 14 m/s, East, in 3.5 sec. What is a?  Given: v o = 0 v f = 14 m/s t = 3.5 s 

9.  9. An automobile slows from 14.0 m/s, East, to 7 m/s, East, in 2.0 sec. What is the average acceleration?  Givens: v o = 14 m/s v f = 7 m/s t = 2 sec  Note acceleration is in the opposite direction to motion  a= (7 -14) /2 = - 3.5 m/s 2

10.  B. Velocity - Time Graphs  1. How Created  a. On d vs t graph the slope of tangent to the graph is the instantaneous velocity  b. Can use these values, for the slope of the tangents to the curved position-time graph, to generate a plot of velocity versus time.  c. Note: velocity being plotted is at that particular instant in time

11. (1) Straight Line (2) Area under the curve equals displacement –2. Example – a. Plots velocity as a function of time – b. constant velocity - no acceleration

12.  c. Velocity is changing - not constant

13.  3. Remember - the area under the curve on a velocity time graph is displacement  4. Slope of the velocity-time graph is acceleration  a. remember that slope is rise over run, and on the vel vs time graph that becomes change in velocity divided by change in time  b. a =  v /  t average acceleration  c. constant slope, than constant acceleration

14.  d. changing slope, means acceleration is changing, so instantaneous acceleration  (1) draw tangent to graph  (2) slope of tangent is acceleration at that point in time  5. Positive acceleration does not always mean velocity is increasing - it depends on sign of velocity  6. Calculate acceleration by subtracting an initial velocity from a final velocity and dividing by the time interval

15.  Remember a = (v f -v i )/t  Solving for v f gives v f = v i + a t  a. assumes uniform acceleration so velocity is constantly changing  b. can have zero velocity with a constant acceleration  (1) Throw a ball straight up. What happens?  (2) Roll a ball up an inclined plane.  Acceleration of gravity = g = -9.8m/s 2  With respect to upward motion  v f = v i + a tif: v f = 9.8m/s - 9.8m/s 2 (t)  At what time will v f = 0s ??  After 1 second

16. Section 4.2 Displacement During Constant Acceleration Objectives:  Calculate displacement given a uniform (constant) acceleration  Solve problems of objects uniformly accelerated by gravity  Learn an organized strategy to solve motion problems

17. Remember…………  We learned the area under the curve on a velocity time graph is displacement  The slope of the velocity-time graph is acceleration  Slope is rise over run, and on the velocity vs. time graph that becomes the change in velocity divided by change in time  b. a =  v /  t = average acceleration  c. constant slope, than constant acceleration

18. (1) Straight Line (2) Area under the curve equals displacement (3) d = vt – Example – A plot of velocity as a function of time – A constant velocity - no acceleration

19. If an object is moving with uniform acceleration, then you can find displacement: so if d = v t, then 1 2 x x Add the areas and you get d = v i t+ ½(v f -v i )t or d = ½(v f +v i )t  Calculating the area under the curve: t vfvf vivi  This is the displacement when the velocity and time are known

20. Displacement when Acceleration is known along with time Since: v f = v i +a t Since: d = ½(v f +v i )t & v f = v i +a t * Combining the two and simplifying gives the following formula: Displacement of an object at constant velocity added to the displacement of an object with uniform acceleration

21. Displacement when velocity and acceleration are known Since: v f = v i +a t Since: d = ½(v f +v i )t & v f = v i +a t Solving for t and combining to eliminate t gives the following formula:Solving for t and combining to eliminate t gives the following formula: a d = ½(v f 2 - v i 2 )/a Solving for v f 2 yields:Solving for v f 2 yields: ad v f 2 = v i 2 + 2ad

22. Displacement when velocity and acceleration are known Example: A speedboat increases velocity from 20 m/s to 30 m/s in a distance of 200m. Find acceleration and time it took. Givens: v i = 20 m/s, v f = 30 m/s, d = 200m v f 2 = v i 2 + 2ad so…. (30m/s) 2 = (20m/s) 2 + 2a(200m) Solving for a gives: a = (500m 2 /s 2 )/400m = 1.25m / s 2 and since v f = v i + a t or t = (v f -v i )/a Then: t = (10m/s)/1.25m/s 2 = 8s

23. Determining Motion Mathematically  Kinematic equations  These equations should be memorized!

24. Problem-Solving Strategy 1.Draw a picture. 2.List variables. 3.Determine proper equation. 4.Replace variables in equation with known values and their units! 5.Solve. 6.Check to make sure units are correct.

25. Example Problem  An automobile starts at rest and speeds up at 3.5 m/s 2 after the traffic light turns green. How far will it have gone when it is traveling at 25 m/s?  Use:  (25m/s) 2 = 2(3.5m/s 2 )(  d)   d = 625m 2 /s 2 ÷ 7m/s 2 = 89.3m dddd

26. Free Fall & Acceleration Objectives:  Solve problems of objects that are uniformly accelerated by gravity  Show the motion of free falling object using both a velocity-time graph and position time graph

27.  Acceleration due to gravity  The acceleration of gravity is given a special symbol, “g”  Acceleration is a vector quantity so “g” has magnitude and direction.  Average value of “g” for the earth’s surface is - 9.80 m/s 2  “Up” is our positive direction. The velocity of a falling object is negative. So as a ball falls its velocity and acceleration are negative. g = -9.80 m/s 2 Note: Pay special attention to the graphs on page 77 in your text

28. Graphs of free fall

29.  3. Uniform acceleration - so motion is symmetrical  a. Objects under only the influence of gravity exhibit the following motion characteristics - freefall  (1) If object is thrown up or is moving up it takes the same time to rise as to fall  (2) It rises same distance as it falls  (3) The magnitudes of velocity are the same at the end as at the beginning - just different signs  (4) The velocity at the top is zero  http://www.youtube.com/watch?v=quhelbcrl78 http://www.youtube.com/watch?v=quhelbcrl78

30.  Example: A baseball is hit straight up in the air with an initial velocity of 38 m/s.  (1) Why use the phrase ‘straight up’?  (2) How long did it stay in the air?  (3) How high did it go?  Givens: v o = 38 m/s, a=-9.8m/s 2 --At the top of the arc the velocity of the baseball is zero - since gravity is acting. v f = -38m/s

31. Gravity Equations  The same as the kinematic equations

32. Look at pages 76 and 77 in text  Understand the graphs on page 77  read, re-read, and look again at the next to last paragraph - “If you throw a ball straight up...”  Work prob 25 - 26 page 77

33. Chapter 4 Accelerated Motion Review  What is acceleration?  The change in velocity over the time interval in which it occurs  What is constant acceleration?  Uniform acceleration. The acceleration does not change over time.  What is the slope of a velocity time graph?  The acceleration  What type of curve is on a velocity time graph for a uniformly accelerated object?  A straight sloped line.

34. Chapter 4 Accelerated Motion Review  What are the equations called that relate the displacement of a uniformly accelerated object to velocity, time and acceleration?  Kinematic Equations  What is the shape of a position time graph for a uniformly accelerated object?  Half a parabola  What is the acceleration of gravity?  -9.8 m/s 2 (near the Earth’s surface)  How can you solve problems of uniformly accelerated motion?  Using the Kinematic Equations (table 4-2, pg78 in text)  How are the equations to solve gravity problems the same as the kinematic equations?  Just replace a with g!

35. Gravity Equations  The same as the kinematic equations

36. What are the steps for Problem-Solving? 1.Read the problem carefully, visualize and draw a sketch 2.Identify and list the know quantities/variables 3.Identify the unknown (what you need to find) 4.Select the proper equation or equations 5.Re-arrange the equations as needed to solve for the unknown 6.Replace variables in equation with known values and their units! 7.Solve and check to make sure units are correct! 8.Double check your work. Do a rough estimate to see if your answer makes sense.