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Constant Motion Jeopardy

Constant Motion Jeopardy. Part 1: Relative Motion. Relative Motion – 10pts Describe the motion of the middle kayak according to the observer in the back. Picture 1 Picture 2 Picture 3. Answer. Main. Relative Motion Answer – 10pts.

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Constant Motion Jeopardy

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  1. Constant Motion Jeopardy

  2. Part 1: Relative Motion

  3. Relative Motion – 10ptsDescribe the motion of the middle kayak according to the observer in the back. Picture 1 Picture 2 Picture 3 Answer Main

  4. Relative Motion Answer – 10pts • The kayak appears to be moving forwards – away from the observer. Question Main

  5. Relative Motion – 20ptsDescribe the motion of the middle kayak according to the observer in the front. Picture 1 Picture 2 Picture 3 Answer Main

  6. Relative Motion Answer – 20pts • The kayak appears to be moving forwards – towards the observer. Question Main

  7. Relative Motion – 30pts • Choose one observer who would see the person moving backwards. Photo 1 Photo 2 Answer Main

  8. Relative Motion Answer – 30pts • Anything moving faster than the person would see the person moving backwards (ex: the train) Photo 1 Photo 2 Question Main

  9. Relative Motion – 40pts • Choose one observer who would see the person as not moving. Photo 1 Photo 2 Answer Main

  10. Relative Motion Answer – 40pts • Anything moving at the same speed and in the same direction as the person, would see the person as not moving (ex. The backpack) Photo 1 Photo 2 Question Main

  11. Part 2: Position vs Time Graphs Main

  12. Position vs. Time – 10 ptsWhere is the object at 36 seconds? Answer Main

  13. Position vs. Time Answer – 10pts • The object is at -30 m. Question Main

  14. Position vs. Time – 20ptsWhere does the object have a constant velocity? Answer Main

  15. Position vs. Time Answer – 20pts • Constant negative velocity: 0 s – 12 s • Constant negative velocity: 36 s – 40 s  Because the slope is a straight line • Constant velocity of zero: 20 s – 28 s Question Main

  16. Position vs. Time – 30ptsWhere does the object have zero velocity? Answer Main

  17. Position vs. Time Answers – 30pts • The object has a constant velocity of zero over the time interval: 20 s – 28 s  Because the slope is horizontal Question Main

  18. Position vs. Time – 40ptsWhat is the average velocity of the object over the total time? Answer Main

  19. Position vs. Time Answer – 40pts v = ∆x/∆t ∆x = xf – xi = (-50 m) – (30 m) = -80 m ∆t = tf – ti = (40 s) – (0 s) = 40 s v = (-80 m)/(40 s) = -2 m/s Question Main

  20. Part 3: Representing Motion Main

  21. Representing – 10ptsDescribe the motion of each hiker in words. Hiker 1:x = (–0.50 m/s)t +(250 m) Hiker 2:x = (0.75 m/s)t + (-150 m) Answer Main

  22. Representing Answer – 10pts • Hiker 1 starts at 250 m up from a camp on the side of a mountain and is walking down the mountain (in the negative direction) at -0.5 m/s • Hiker 2 starts -150 m below the camp on the side of the mountain and is walking up the mountain (in the positive direction) at 0.75 m/s Question Main

  23. Representing – 20ptsDraw a picture representing the motion of each hiker. Hiker 1:x = (–0.50 m/s)t +(250 m) Hiker 2:x= (0.75 m/s)t + (-150 m) Answer Main

  24. + Representing Answer – 20 pts Hiker 1 Camp at 0 m Hiker 2 – Question Main

  25. Representing – 30ptsDraw a dot diagram representing the motion of each hiker. Hiker 1:x = (–0.50 m/s)t +(250 m) Hiker 2:x = (0.75 m/s)t + (-150 m) Answer Main

  26. Representing Answer – 30pts + Camp at 0 m - . . . . . . . . 0s 1s 2s 3s 4s 5s 6s 7s Hiker 1 . . . . . . 5s 4s 3s 2s 1s 0s Hiker 2 Question Main

  27. DOUBLE JEOPARDY Representing – 40ptsPlot a Position vs. Time graph for Hiker 1 and Hiker 2’s motion Hiker 1:x = (–0.50 m/s)t +(250 m) Hiker 2:x = (0.75 m/s)t + (-150 m) Answer Main

  28. Representing Answer – 40pts Question Main

  29. Part 4: Working with equations Main

  30. Equations – 10ptsWhere will Hiker 1 be after 300 s? Hiker 1:x = (–0.50 m/s)t +(250 m) Hiker 2:x = (0.75 m/s)t + (-150 m) Answer Main

  31. Equations Answer – 10pts • Hiker 1: x = (–0.50 m/s)t +(250 m) t = 300 s x = (–0.50 m/s)(300 s)+(250 m) x = (-150 m) + (250 m) x = 100 m Question Main

  32. Equations – 20ptsWhere will Hiker 2 be after 300 s? Hiker 1:x = (–0.50 m/s)t +(250 m) Hiker 2:x = (0.75 m/s)t + (-150 m) Answer Main

  33. Equations Answer – 20pts • Hiker 2: x = (0.75 m/s)t + (-150 m) t = 300 s x = (0.75 m/s)(300 s) + (-150 m) x = (225 m) + (-150 m) x = 75 m Question Main

  34. Equations – 30ptsWrite an equation for the motion of the object below. Answer Main

  35. Equations Answer – 30pts xi = 15 m v = ∆x/∆t = (-20 m)/(5 s) = -4 m/s ∆x = (-5 m) – (15 m) = -20 m ∆t = (5 s) – (0 s) = 5 s x = (-4 m/s)t + (15 m) Question Main

  36. Equations – 40ptsWrite an equation for the motion of the object below. Answer Main

  37. Equations Answer – 40pts xi = 6 m v = -3 m/s x = (-3 m/s)t + (6 m) Question Main

  38. Physics CP Jeopardy Kinematics: Constant Motion Review Main

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