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Distinguish displacement from distance. Distance is a scalar quantity, which means that is only needs a magnitude. It is how far you are from a reference point.Displacement is a vector quantity, which means it needs both magnitude and direction. It is how far you ended up from where you started and in what direction. Therefore, distance is the magnitude of a displacement vector. .

Linear Motion

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**1. **Linear Motion Holt Physics, pages 39 - 81

**2. **Distinguish displacement from distance Distance is a scalar quantity, which means that is only needs a magnitude. It is how far you are from a reference point.
Displacement is a vector quantity, which means it needs both magnitude and direction. It is how far you ended up from where you started and in what direction. Therefore, distance is the magnitude of a displacement vector.

**3. **Distinguish displacement from distance (cont’d) For instance, if you leave home and drive from Houston to Dallas, your displacement is approximately 250 mi, North. When you return to Houston at the end of the trip, your displacement is now zero because you are now back at the place you started.

**4. **Distinguish velocity from speed Speed is a scalar quantity. It is how fast you are traveling. When you look at the speedometer on your car, it tells you the speed you are traveling at that moment.
Velocity is a vector quantity. It is how fast you are going and in what direction. If you are traveling to Dallas, you may be traveling at 70 mi/hr, North.

**5. **Define frame of reference. A frame of reference is necessary to fully explain motion. It is the position from which the motion is being observed.
The frame of reference is assumed to be stationary.
Your frame of reference affects the motion you perceive.

**6. **Give an example of a frame of reference. If you are looking at Spring from across the street, it appears to be sitting still but if you were viewing it though a telescope from the moon, it would appear to be rotating along with the earth which is turning on it’s axis.

**7. **Define average velocity Average velocity is the total change in position divided by the time interval over which it occurred.

**8. **Sample Problem Suppose a car travels at a constant 10 m/s. How far would it move in one minute?

**9. **Sample Problem You drive a car 2 hours at 40 km/h, then 2 hours at 60 km/h. What is your average velocity? Do you get the same answer if you drive 100 km at each of these two speeds?

**10. **Define the concept of relativity of velocities. Velocity is also affected by the frame of reference. This is known as relative velocity

**11. **Use the concept of relativity of velocities. (cont’d) If you are riding on the bus talking to your friend, his apparent speed is 0 m/s.
However if you are standing on the street corner and the bus drives by without stopping, his apparent speed is now 40 mi/hr (the speed of the bus).
If you are in your mom’s car going 50 mi/hr when you pass your friend on the bus that is going 40 mi/hr then your friend appears to be going –10 mi/hr (or 10 mi/hr in the opposite direction).

**12. **Define acceleration Acceleration is a vector quantity. It is the rate at which the velocity is changing. Since velocity is dependent on either speed or direction, if either of those change, your velocity changes and you have acceleration.
Deceleration is a special name given to an acceleration that is in a direction opposite to the direction of motion of the object.

**13. **Acceleration equations Since velocity, displacement, and acceleration are all vector quantities, you must keep direction in mind when you substitute values into the equations. Up or to the right are considered to be positive directions. Down or to the left are considered to be negative directions.

**14. **Sample Problem It takes 4.8 seconds for a car’s speed to increase by 10 m/s. What is its acceleration?

**15. **Acceleration Equations a – acceleration in m/s2
?v - change in velocity in m/s
vf – final velocity in m/s
vi – initial velocity in m/s
?t or t – time interval in seconds
d – displacement in m

**16. **Sample Problem A rocket is capable of accelerating at 800 m/s2. How long after lift off will the rocket reach 500 m/s?

**17. **Sample Problem A car is traveling at 50 m/s must slow down to 30 m/s in the next 10 m. What deceleration must the car have?

**18. **Sample Problem A car moves at 12 m/s and coasts up a hill with a uniform acceleration of -1.6 m/s2. How far has it traveled after 6 sec? How far has it gone after 9 sec?

**19. **Sample Problem An engineer must design a runway to accommodate airplanes that must reach a ground velocity of 61 m/s before they take off. These planes are capable of being accelerated uniformly at the rate of 2.5 m/s2. How long will it take the planes to reach takeoff speed? What must be the minimum length of the runway?

**20. **Sample Problem As a traffic light turns green, a waiting car starts with a constant acceleration of 6 m/s2. At the instant the car begins to accelerate, a truck with a constant velocity of 21 m/s passes in the next lane. How far will the car travel before it overtakes the truck? How fast will the car be traveling when it overtakes the truck? HINT: Set the two distances equations equal to each other.

**21. **Plot and interpret position-time graphs A position time graph shows an objects change in position over a period of time.

**22. **Plot and interpret position-time graphs An object which is not moving would have the same position over the period of time, so the graph would be a horizontal line. The y-intercept of this line indicates the distance of the object from a stationary reference point.

**23. **Plot and interpret position-time graphs (cont’d) An object which is moving at a constant positive velocity would cover equal distances in equal amount of time and it’s graph would appear as an upwardly sloping diagonal line.

**24. **Plot and interpret position-time graphs (cont’d) An object which is moving with a constant negative velocity would also cover equal distances in equal amounts of time but its distance from the reference point would be decreasing. Its graph would be a downward sloping diagonal line.

**25. **Plot and interpret position-time graphs (cont’d) An object which is moving with a constant positive acceleration will cover larger and larger distances in successive equal time intervals. Its position time graph would appear as a curve.

**26. **Plot and interpret position-time graphs (cont’d) An object which is moving with a constant negative acceleration will cover smaller and smaller distances in successive equal time intervals.

**27. **Calculate the velocity of an object from a position time graph The slope of the position time graph is the velocity of the object.

**28. **Calculate the velocity of an object from a position time graph

**29. **Plot and interpret position-time graphs The faster an object is moving, the steeper the line on the position time graph.

**30. **Define instantaneous velocity Instantaneous velocity is the velocity at any instant.
When an object is moving at a constant velocity, the instantaneous velocity and the average velocity are the same.
When an object is accelerating, the instantaneous velocity and the average velocity are not the same.

**31. **Define instantaneous velocity (cont’d) To find the instantaneous velocity of an accelerating object from a position time graph, you can find the slope of the line that is tangent to the curve at the point in time for which you want the instantaneous velocity.

**32. **Plot and interpret a velocity-time graph An object which is not moving would have no velocity over the entire period of time, so the graph would be a horizontal line along the x-axis.

**33. **Plot and interpret a velocity-time graph An object which is moving at a constant positive velocity would maintain the same velocity over the entire time interval and it’s graph would appear as a horizontal line above the x-axis. The y-intercept indicates the constant velocity of the object.

**34. **Plot and interpret a velocity-time graph An object which is moving at a constant negative velocity would maintain the same velocity over the entire time interval and it’s graph would appear as a horizontal line below the x-axis. The y-intercept indicates the constant velocity of the object.

**35. **Plot and interpret a velocity-time graph An object which is moving with a constant positive acceleration will have a velocity time graph that is an upwardly sloping diagonal line. The y-intercept is the initial velocity of the object.

**36. **Plot and interpret a velocity-time graph An object which is moving with a constant negative acceleration will have a velocity time graph that is an downwardly sloping diagonal line. The y-intercept is the initial velocity of the object.

**37. **Calculate the acceleration of an object from a velocity time graph The slope of the velocity time graph is the acceleration of the object.

**38. **Calculate the acceleration of an object from a velocity time graph

**39. **Calculate the displacement from a velocity time graph The area between the velocity time graph and the x-axis is the displacement of the object during that time interval.

**40. **Sample Problem What is the displacement of the object during the first 5 seconds?

**41. **Sample Problem (cont’d)

**42. **Sample Problem (cont’d)

**43. **Sample Problem (cont’d) To find the total displacement of the object during the 5 seconds, you would break the shape into easily calculated areas as shown, calculate the individual areas then add them together.

**44. **Plot and interpret an acceleration-time graph An object which is not moving would have no acceleration over the entire period of time, so the graph would be a horizontal line along the x-axis.

**45. **Plot and interpret an acceleration-time graph An object which is moving with either a constant positive or negative velocity would have no acceleration over the entire period of time since the velocity is not changing. So the graph would be a horizontal line along the x-axis.

**46. **Plot and interpret an acceleration-time graph An object which is moving with constant positive acceleration would graph as a horizontal line above the x-axis. The y-intercept would be equal to the value of the acceleration.

**47. **Plot and interpret an acceleration-time graph An object which is moving with constant negative acceleration would graph as a horizontal line below the x-axis. The y-intercept would be equal to the value of the acceleration.

**48. **Calculate the velocity from an acceleration time graph The area between the acceleration time graph and the x-axis is the velocity of the object.

**49. **Sample Problem What is the velocity of the object after 10 seconds?

**50. **Sample Problem (cont’d) To calculate the velocity, you would find the area of the shaded region.

**51. **Sample Problem (cont’d)

**52. **Constant Positive Velocity

**53. **Acceleration due to gravity If an object is dropped or thrown, it will fall under the acceleration of gravity and it’s acceleration is 9.8 m/s2, down. The entire time the object is in the air, it has an acceleration of 9.8 m/s2, down, but its velocity is constantly changing. On the way upward, the speed is decreasing and on the way down, the speed is increasing.

**54. **Acceleration due to Gravity Keep in mind that acceleration is a vector quantity so when it is used in an equation, you use a + or - sign to indicate whether the direction is up or down. Since the acceleration due to gravity is 9.8 m/s2, down when it is substituted into the equations you will use -9.8 m/s2.

**55. **Gravity equations You will use the same equations however, you will use the value of -9.8 m/s2 for the acceleration when an object is in freefall.

**56. **Sample Problem A stone falls freely from rest for 8 seconds. What is the stone’s velocity after 8 seconds? What is the stone’s displacement during this time?

**57. **Sample Problem Kyle is flying a helicopter that is rising at 5 m/s when he releases a bag. After 2 seconds, what is the bag’s velocity? How far has the bag fallen? How far below the helicopter is the bag?