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Motion in One Dimension

Motion in One Dimension. Chapter 2. Physics Lingo. Dimension: the minimum number of coordinates needed to specify the points within (NSEW, Quadrants of a graph) Magnitude: the size of something Scalar: fully described by a number/magnitude alone

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Motion in One Dimension

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  1. Motion in One Dimension Chapter 2

  2. Physics Lingo • Dimension: the minimum number of coordinates needed to specify the points within (NSEW, Quadrants of a graph) • Magnitude: the size of something • Scalar: fully described by a number/magnitude alone • Vector: described by BOTH a magnitude AND direction

  3. Pythagorean Theorem • Pythagorean Theorem: used to find the missing length of a right triangle

  4. Trigonometric Functions

  5. Physics Chapter 2 • Day 1: Displacement and velocity • Day 2: Acceleration • Day 3: Graphing motion activity • Day 4: Free fall • Day 5: Labs • Day 6: Review • Day 7: Test

  6. 1D Motion • 1D is simplest form • One direction (straight track) • Forward or Backward, up or down • Motion: • Takes place over time • Has a frame of reference • Provides a starting point • Set the starting point/zero point • Do not move it • Helps to determine directional signs

  7. Change in position • Change in position calculations • Horizontal (x= xf-xi) • Vertical (y= yf-yi) • Could be 0 if returns to starting point • Give direction AND number AND unit • “+” if right/east/up • “-” if left/west/down

  8. Displacement vs. distance • Definitions: • Distance: sum of all segments regardless of direction • Displacement: where you end up in relation to start • Examples: • Walk 5 m right then 10 m left. Distance? Displacement? • Walk 12 m left then 12 m right. • Walk a mile circular track.

  9. Velocity • Speed has no direction • Velocity has magnitude AND direction • Units: m/s • The sign tells direction • The number represents magnitude

  10. Go To Grandma’s House • Use polar coordinates (NSEW) to draw a map to “Grandma’s House” from WHHS. Have at least 4 turns. • Find the Distance traveled. • Find the Displacement. • Pretend it took 35 minutes to get there. • Determine your speed. • Determine your velocity. • Predict what type of vehicle you used to travel.

  11. Graphing Motion • Plot Position vs. Time • X-axis is Time (s) • Y-axis is Position/Distance (m)

  12. Types of Velocity Graphs • Straight line: • Constant velocity • y= mx+b and m=velocity because

  13. Types of Graphs • Curved line: • Varied velocity • Find instantaneous velocity • Draw a tangent line • Slope of tangent line = instantaneous velocity

  14. Interpreting Velocity Graphs

  15. Interpreting Velocity Graphs

  16. Interpreting velocity graphs

  17. Homework • Pg. 47-48 • 1, 7, 17

  18. Acceleration • Acceleration: change of velocity (magnitude or direction) in a certain time interval • Units: m/s2 • v=“+” then the object is speeding up • v=0 then the object is at a constant speed • v=“-” then the object is slowing down

  19. Acceleration VS. Velocity • What would the sign be for velocity? Acceleration?

  20. Graphing Motion • Distance vs. Time Graphs • What does this imply about acceleration? • +a, -a, a=0?

  21. Graphing Acceleration • Plot Velocity vs. Time • X-axis is Time (s) • Y-axis is Velocity (m/s)

  22. Graphing Acceleration • What do these graphs tell you about acceleration?

  23. Interpreting Graphs • Determine the acceleration for A, B, and C segments in each graph.

  24. Interpreting Graphs • Which color graph represents the puck’s motion from (a)?

  25. Interpreting Graphs • Find the average acceleration of the object during the time intervals 0 to 5.0 s, 5.0 s to 15 s, and 0 to 20 s.

  26. Classwork and HW • Tickertape lab • HW: • Pg. 46-47 • 1,2,3,9,14

  27. Equations for Acceleration • Equations for 1D Motion with Constant Accelerated (in packet, no “x”) +)t (no “a”) x=vot + at2(in packet, no final v) v2=vo2+2ax (in packet, no “t”)

  28. Using Motion Equations • Read the problem. • Draw a diagram (use arrows to show direction and indicate initials and finals) • Label all quantities. • Choose appropriate equation(s). • Solve for unknowns. a= 5.00 m/s2

  29. Using Motion Equations • A car traveling at a constant speed of 24.0 m/s passes a trooper hidden behind a sign. One second after the car passes the sign, the trooper sets off in chase with a constant acceleration of 3.00 m/s2. How long does it take for the trooper to overtake the car? How fast is the trooper going at that time? • Draw a diagram. • Label quantities. • Choose appropriate equations. • Solve.

  30. Using Motion Equations • How long does it take for the trooper to overtake the car? • If t=0 is when the trooper starts to chase, the car would have reached 24.0 m. When the trooper catches the car, xcar=xtrooper. • Xcar= xo+vot+(1/2)at2= 24.0m + (24.0 m/s)t + 0 • xtrooper=xo+vot+(1/2)at2 = 0 + 0 + (1/2) (3.00 m/s2)t2 • Set these two equations equal to each other. • How fast is the trooper going at that time? • Substitute the time from above into a velocity equation. • Vtrooper= vo + atroopert

  31. Using Motion Equations • A typical jetliner lands at a speed of 71.5 m/s and decelerates at the rate of 4.46 m/s2. If the plane travels at a constant speed for 1.0 s after landing and before applying the brakes, what is the total displacement of the aircraft before coming to rest? x=vot + (1/2)at2 V2=v02 + 2a(x-xo)

  32. Classwork and HW • Graphing Motion Activity • HW • Pg. 49-50 • 19, 30, 37, 39

  33. Freely Falling Objects • Free fall: any object moving freely under the influence of gravity alone, regardless of its initial motion (includes objects thrown upward, downward or released from rest) • 1 D motion with constant acceleration due to gravity • Objects fall at the same rate • ag (in general) and g on Earth (g=-9.81 m/s2 or -32 ft/s2) • It is -1.62 m/s2 on the Moon. • In a vacuum: • Does a feather or brick hit first when dropped?

  34. “g” always equals -9.81 m/s2 • Upward: it slows down so “-v” • Peak: v=0 • Downward: it speeds up but in the negative direction so “-v” • Rules of Symmetry • tup=tdown • xup=xdown • vrelease=vreturn

  35. Using Motion Equations • A stone is thrown from the top of a building with an initial velocity of 20.0 m/s straight up at an initial height of 50.0m. • What is the time it takes for the stone to reach its maximum height? • What is the maximum height that the stone reaches? • What is the time it takes for the stone to return to the height from which it was thrown? What is the velocity at this instant? v=vo+at y=vot +(1/2)at2 Vrelease=vreturn

  36. Using Motion Equations • This same stone just misses the edge of the roof on its way down. • What is the time it takes for the stone to reach the ground? • yo=50.0 m and yf = 0 m • a=-9.8m/s2 and vo=20 m/s • What is the velocity and position of the stone at after 5.00 s? • v=vo+at • y=yo+vot+(1/2)at2

  37. Classwork and HW • Reaction time lab • HW • Pg. 47 (16) • Pg. 50 (43, 46)

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