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Chapter 2: Describing Motion in 1-D

Chapter 2: Describing Motion in 1-D. Frame of Reference. Whether or not you are moving depends on your point-of-view. From inside the box car, the woman in red is stationary (at rest). From outside the box car, the woman in red is moving.

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Chapter 2: Describing Motion in 1-D

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  1. Chapter 2: Describing Motion in 1-D

  2. Frame of Reference Whether or not you are moving depends on your point-of-view. From inside the box car, the woman in red is stationary (at rest). From outside the box car, the woman in red is moving. So what is it? Is she moving? Is she stationary?

  3. Frame of Reference Here’s a “thought experiment” B A 26 m/s 20 m/s Two trains travel on parallel tracks past a platform. If you are standing on the platform, how fast does each travel? If you are on train A, how fast is train B travelling? If you are on train A, how fast are the other passengers on train A moving relative to YOU?

  4. Frame of Reference General Rule: Any measurement of position, distance, or speed must be made with respect to a reference frame. Unless otherwise stated, the earth will be used as the reference frame.

  5. Distance / Displacement Distance: how far you travelled in total. Displacement: the change in distance. Distance is the total length (measured in meters) of travel along a path. Displacement is the change in position (measured in meters). It depends on the starting and ending points – not the path of travel.

  6. Distance / Displacement Example: You walk 4 blocks North. What is your distance? What is your displacement? Example: You walk 4 blocks North then 7 blocks South. What is your distance? What is your displacement? Example: You walk 4 blocks North, then 7 blocks South, then 3 blocks North. What is your distance? What is your displacement?

  7. Displacement X1 X2 Displacement = ∆X = X2 – X1 Initial position Final position Displacement ONLY depends on the end points, NOT the path taken!

  8. Direction Remember this : • Distance does NOT have a direction. • Displacement DOES have direction. On a number line (or a coordinate system) you can have positive or negative displacement y x Coordinate Plane

  9. Scalars & Vectors • If a quantity has only a numerical value, it is a SCALAR. (it has magnitude) If a quantity has both a numerical value AND a direction associated with it, it is a VECTOR. (it has magnitude and direction) Distance Displacement

  10. How fast are you going? Speed is the distance traveled per time spent to travel. It is a Scalar whose units are m/s. S = d / t There are 2 types of speed we need to consider: Average Speed – Average speed is the total distance traveled divided by the total time it took to travel. It is the average speed of travel. Instantaneous Speed – Instantaneous speed is the speed at a particular moment in time.

  11. And WHERE are you going? Velocity is the change in position (displacement) per time spent to travel. It is a Vector whose units are m/s. V = ∆x / ∆t Velocity is a vector that is based on displacement, so you can change your velocity by changing your speed OR by changing your direction. The units for velocity are m/s

  12. Plotting Position vs. Time The basic Motion plot we will study in Physics is: Position vs Time It looks like this: The slope of a Position vs Time plot will give you the VELOCITY (X2,Y2) Position (m) (X1,Y1) Time (s)

  13. If the plot gives a linear graph, this tells you the object travelled with constant speed. If the plot gives a non-linear graph, then the speed is changing. Increasing speed (the slope keeps increasing) Position (m) Constant speed (the slope is constant) Time (s)

  14. Calculating Speed from a Time vs. Position Graph For objects moving with constant speed, the slope of the plot can tell you much information: Straight line = constant speed Medium slope = medium speed Straight line = constant speed Low slope = low speed Position (m) Position (m) Time (s) Time (s) Straight line = constant speed High slope = high speed Straight line = constant speed No slope = no speed (at rest) Position (m) Position (m) Time (s) Time (s) Velocity animation

  15. Calculating Velocity from a Time vs. Position Graph Remember: Velocity is Speed & Direction. It is a Vector Straight line = constant speed Positive slope = going away from you Position (m) Time (s) Straight line = constant speed Negative slope = coming back toward you Position (m) Time (s) In Physics, the use of a negative sign shows direction Slope tutorial

  16. Onto the next level . . . • When graphically analyzing the motion of an object, start with a time vs. position graph and see what type of motion it is. Look at the slope to gain more detailed info. • Next step: do a time versus velocity graph. The slope of such a graph will give you the ACCELERATION of the object. Acceleration - how much the velocity changes in a specified time interval.

  17. Describe what is happening to the object. What is happening in section A? What is happening in section B? What is happening in section C? Velocity (m/s) 30 25 20 15 10 5 0 B A C 0 3 6 9 12 15 Time (s) Acceleration is the slope of the graph (on a Time vs. Velocity Graph)

  18. Answers A: Object has increasing velocity (speeding up) Object has constant acceleration (straight line slope) a = (30-0)/(9-0) = 3.33 m/s2 B: Object is travelling with constant velocity. Object has no acceleration (no slope) a = (30-30)/(12-6) = 0.00 m/s2 C: Object has decreasing velocity (slowing down) Object has constant acceleration (deceleration) a = (0-30)/(15-12) = -10 m/s2 Velocity (m/s) 30 25 20 15 10 5 0 B A C 0 3 6 9 12 15 Time (s) 9/19/11

  19. Acceleration • Acceleration is a VECTOR – it has magnitude and direction • Acceleration is the slope of a time vs. velocity graph. • Acceleration is the change of velocity in a specified time interval. • a = ∆v / ∆t where: ∆v = vfinal – vinitial ∆t = tfinal – tinitial • The symbol for Acceleration is: a • The units for Acceleration are: m/s2

  20. One last thing . . . . When looking at Velocity vs. Time Graphs, if you want to calculate the object’s displacement (change in position), calculate the area under the graph. For example: Velocity (m/s) The area under the line will give you the object’s displacement in the time interval specified. Displacement = area under line Displacement = ∆v * ∆t ∆v Time (s) ∆t Displacement tutorial

  21. OK, let’s talk a little more about acceleration . . . • We already know that acceleration is how fast your velocity changes. acceleration is the rate of change of velocity. acceleration is the slope of the graph on a time vs. velocity graph Let's see how velocity and acceleration at related for objects in motion: 1. If the object is moving away, it has positive velocity. 2. If the object is moving at you, it has negative velocity. 3. If the object is speeding up, it has positive acceleration. 4. If he object is slowing down it has negative acceleration.

  22. What would be happening if an object has positive velocity and positive acceleration? What would be happening if an object has negative velocity and positive acceleration? What would happen if an object has positive velocity and negative acceleration? What would happen if an object has negative velocity and negative acceleration? moving away from you and increasing speed moving at from you and increasing speed moving away from you and decreasing speed moving at from you and decreasing speed

  23. We know that in our lives, a lot of forces can cause objects to accelerate or decelerate (negative acceleration) • In the lab, you rolled a cart down an inclined plane and saw that it accelerated due to gravity. You calculated its acceleration from the plot of a time vs. velocity graph. • As the angle of inclination of the plane increases, the cart will experience greater and greater acceleration. (it speeds up faster – the rate of change of velocity) • The greatest value of acceleration that an object can have without outside forces on it is 9.81 m/s2. This is the value that gravity will accelerate objects when they fall on their own in a vacuum. This is called: free fall

  24. Free Fall Galileo dropped two balls of different mass from the Leaning Tower of Pisa and found that both landed at the same time. He proved that gravity accelerates all objects to the ground at the same rate. Based on his calculations, when all objects fall freely, they accelerate at 9.81 m/s2 or about 32 ft/s2. (neglecting air resistance)

  25. This means that for every second that an object falls freely, its velocity increases by 9.81 m/s. Here are the graphical representations of an object in freefall: position velocity Power graph = Linear graph = constant acceleration Increasing velocity Slope of line = a = 9.81 m/s2 time time

  26. Let's look at data analysis of an object initially at rest (not moving) and dropped freely.

  27. Freefall can also be considered to be when you throw an object straight up into the air. Think about this: If you drop a ball, it starts out with zero velocity (at rest) and has velocity at the end of its travel. If you throw a ball up into the air, it starts out with velocity you give it and has zero velocity at the end of its travel. They are the same, just opposite directions !!!!!

  28. Analyzing Motion via Equations (solving Physics problems) So far, we have solved motion problems by graphical means. For example: solve the following problem: A car travels 24m in 3s. What is its velocity? You would have plotted this out on a time vs. position graph and found the slope of the line. 24 slope = ∆y = (24-0) = 8m/s ∆x (3-0) Position (m) 3 Time (s)

  29. Now let’s solve it with equations we know . . . How about v = ∆x ∆t So, solve the following problem: A car travels 24m in 3s. What is its velocity? Use the GUESS Method . . . Gwrite your givens. U write your unknown(s). Ewrite the equation you will use. Ssubstitute the given values into the equation. Ssolve the equation (using correct units!) (and I HIGHLY recommend drawing it also!) given: ∆x = 24m ∆t = 3s v = ? v = ∆x / ∆t v = 24m / 3s V = 8m/s

  30. Analyzing Freefall via Equations (solving Physics problems) OK, we now know that when we drop an object, it will increase its velocity as it falls. Let’s try this freefall problem: A ball is dropped off a cliff. It falls for 4s and accelerates at 9.81 m/s2. How fast is it going? Now let’s take that information and apply it to find out more about it’s fall . . . How far did it travel? Make sure you use GUESS !!!!!!

  31. Kinematic Equations Equations we already know: Speed Velocity Acceleration S = d / t v = ∆d / ∆t a = ∆v / ∆t and: ∆v = vfinal– vinitial∆t = tfinal – tinitialvavg = (vfinal + vinitial) 2 Let’s rewrite the acceleration equation: Start with a = ∆v / ∆t Substitute in for ∆v . . . ∆v = vfinal– vinitial Multiple both sides by ∆t Add vinitial to both sides Write in format: vfinal = a = ∆v / ∆t a = (vfinal– vinitial) / ∆t a ∆t = (vfinal – vinitial) a ∆t + vinitial = vfinal vfinal= vinitial + a ∆t

  32. This is the 1st Kinematic Equation: vf = vi + a ∆t

  33. Speed Velocity Acceleration 1st Kinematic S = d / t v = ∆d / ∆t a = ∆v / ∆t vf = vi + a ∆t and: ∆v = vfinal – vinitial∆t = tfinal – tinitialvavg = (vfinal + vinitial) 2 Let’s rewrite some of the above equations: Start with v = ∆d / ∆t And we have vavg = (vfinal + vinitial) / 2 If there is constant acceleration, v =vavg Rewrite it in ∆d = format Substitute for vfinal . . . vf = vi + a ∆t Clean up the equation v = ∆d / ∆t vavg = (vfinal + vinitial) / 2 (vfinal + vinitial) / 2 = ∆d / ∆t ∆d = (1/2) (vfinal + vinitial) * ∆t ∆d = (1/2) (vinitial + a ∆t + vinitial) * ∆t ∆d = (1/2) (a ∆t2) + (vinitial) * ∆t

  34. This is the 2nd Kinematic Equation: ∆d = (vi) ∆t + 1/2(a ∆t2)

  35. Speed Velocity Acceleration 1st Kinematic 2nd Kinematic S = d / t v = ∆d / ∆t a = ∆v / ∆t vf = vi + a ∆t ∆d = (vi) ∆t + 1/2(a ∆t2) and: ∆v = vfinal – vinitial∆t = tfinal – tinitialvavg = (vfinal + vinitial) 2 Let’s rewrite some of the above equations: Start with the 1st Kinematic Equation Bring vi to the left side Substitute for ∆t . . . v = ∆d / ∆t If there is constant acceleration, v = vavg Multiply both sides by (vf + vi) Multiply out the left side of the equation Rewrite the equation vf = vi + a ∆t vf - vi = a ∆t vf - vi = a (∆d / v) vf - vi = a ∆d / (vf + vi) /2 (vf + vi) (vf – vi)= 2 a ∆d vf2 – vi2= 2 a ∆d vf2 = vi2 + 2 a ∆d

  36. This is the 3rd Kinematic Equation: vf2 = vi2 + 2 a ∆d

  37. Let’s take a closer look at FreeFall motion so we can really use the Kinematic Equations to start solving some problems: Let’s look at Going down? Going up? • If dropped, an object’s initial velocity is 0m/s • And you will be given or need to calculate its final velocity. • If thrown downward, then you need to make this its initial velocity. • Acceleration due to gravity is 9.81m/s2 and is positive because the object is speeding up as it falls. • If thrown upward, an object’s final velocity is 0m/s at the top of its flight. • And you will be given or need to calculate its initial velocity. • If thrown upward, then you need to make this its initial velocity. • Acceleration due to gravity is 9.81m/s2 and is negative because the object is slowing down up as it goes upward.

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