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Speed or Velocity Time Graphs

Speed or Velocity Time Graphs. Do In Notes: Sketch a d – t graph for object moving at constant speed. Now sketch a speed time graph showing the same motion. d. v. t. t. Constant Velocity/Speed. What is an object is uniformly speeding up?.

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Speed or Velocity Time Graphs

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  1. Speed or Velocity Time Graphs

  2. Do In Notes: Sketch a d – t graph for object moving at constant speed.Now sketch a speed time graph showing the same motion. d v t t

  3. Constant Velocity/Speed

  4. What is an object is uniformly speeding up?

  5. Sketch a distance – t graph for object starting from rest and speeding up with constant acceleration.Now sketch a speed time graph showing the same motion. v d t t

  6. Constant / Uniform Acceleration.Slope = Constant Acceleration of straight line on v-t graph. Average Velocity is the midpoint between 2 speeds. vf + vi 2

  7. What is the acceleration of this object? • Slope = accl. Dy/ Dx • (50 – 10) m/s • (5 – 1) s • What is the sign of accl? • What is the average speed between 1 – 3 seconds? • What is the average speed between 3 – 5 seconds?

  8. . What is the acceleration between:0 – 3 seconds, 5-10 seconds? • Slope = 2 m/s2. • 0.

  9. What’s going on here?

  10. Velocity Time GraphsVector Nature sign of acceleration

  11. Sign of velocity.

  12. Finding Distance or Displacement on V-T graphsSpeed Time = distanceVelocity Time = displacement

  13. Displacement = Area Under Curve at specific time. Drop a vertical to the X axis. What is the displacement at 20 sec? A = bh = (20s)(30m/s)

  14. 2. What is the displacement at 5 s? Area = bh = (5 s) x (1 m/s) = 5 m.

  15. 3: What is the displacement at 4 seconds? A = ½ bh = ½ (4s)(40 m/s) = 80 m.

  16. 4. What is the displacement at 10 s? • A1 = 1/2bh = 1/2(4s)(8m/s) = 16 m • A2 = bh = (6 s) (8 m/s) = 48 m • A tot = 64 m.

  17. Return to start point

  18. How can you tell when object is back to starting point? • Positive displacement = negative displacement. • Tot displacement = 0.

  19. 5. At what time does the object return to the starting point? • At 5 seconds d = 5 m. • From 5 – 10 seconds d = - 5 m. • At t = 10 s.

  20. Given the v – t graph below, sketch the acceleration – t graph for the same motion.

  21. Acceleration – time Graphs • What is the physical behavior of the object? • Slowing down pos direction, constant vel neg accel.

  22. d-t: • slope = velocity • area ≠ . • v-t: • slope = accl • area = displ • a-t: • slope ≠ . • area = D vel • vf – vi.

  23. Hwk. Text do pg 70 #17, 30 and packet “Motion Graph Prac”.In class pg 59 #5.

  24. Objects Falling Under Gravity

  25. Freefall Gravity accelerates uniformly masses as they fall and rise. Earth’s acceleration rate is 9.81 m/s2 – very close to 10 m/s2.

  26. Falling objects accelerate at the same rate in absence of air resistance

  27. But with air resistance

  28. Fortunately there is a “terminal fall velocity.” After a while, the diver falls with constant velocity due to air resistance. Unfortunately terminal fall velocity is too large to live through the drop.

  29. Apparent WeightlessnessObjects in Free-fall Feel Weightless

  30. What is the graph of a ball dropped?

  31. What do the d-t, v-t, and a – t, graphs of a ball thrown into the air look like if it is caught at the same height?

  32. A ball is thrown upward from the ground level returns to same height. a is -9.81, the ball is accelerating at constant 9.81 m/s2. velocity is + when the ball is moving upward d = ball’s height above the ground Why is acceleration negative? Is there ever deceleration?

  33. Free-fall Assumptions Trip only in the air. Trip ends before ball caught. -Symmetrical Trip time up = time down -Top of arc: v = 0, a = ?? -On Earth g = -9.81 m/s2. Other planets g is different.

  34. List given quantities & unknown quantity. • Choose accl equation that includes known & 1 unknown quantity. • Be consistent with units & signs. • Check that the answer seems reasonable • Remain calm Solving:Use accl equations replace a with -g.

  35. Practice Problem. • 1. A ball is tossed upward into the air from the edge of a cliff with a velocity of 25 m/s. It stays airborne for 5 seconds. What is its total displacement?

  36. vi = +25 m/sa = g = -9.81 m/s2. t = 5 s.d = ? • d= vit + ½ at2. • (25m/s)(5s) + 1/2(-9.81 m/s2)(5 s)2. • 125 m - 122 . 6 = +2.4m. • It is 2.4m above the start point.

  37. 2. If the air time from the previous problem is increased to 5.2 seconds, what will be the displacement? • d = vit + ½ at2. • -2.6 m • It will be below the start point.

  38. Ex 3. A 10-kg rock is dropped from a 7- m cliff. What is its velocity just before hitting the ground? • d = 7m • a = -9.81 m/s2. • vf = ? • Hmmm • vi = 0. • vf2 = vi2 + 2ad • vf2 = 2(-9.81m/s2)(7 m) • vf = -11.7 m/s (down)

  39. 4. A ball is thrown straight up into the air with a velocity of 25 m/s. Create a table showing the balls position, velocity, and acceleration for each second for the first 5 seconds of its motion. •  T(s) d v (m/s) a (m/s2) • 0 0 25 -9.81 • 1 20 15.2 -9.81 • 2 30 5.4 -9.81 • 3 31 -4.43 -9.81 • 5 2.4 -24 -9.91

  40. Read Text pg 60-64 Do prb’s pg 64# 1-5 show work.

  41. Mech Universe: The Law of Falling Bodies: http://www.learner.org/resources/series42.html?pop=yes&pid=549#

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