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Ch5 – Mathematical Models

Ch5 – Mathematical Models. 30 25 dist (m) 20 15 10 5. Time (sec) Position (m) 0 10 1 12 2 14 3 16 4 20 5 26. Where is the runner: at 2 sec? at 3 sec? at 4½ sec? At what time was the runner at the 15m mark?

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Ch5 – Mathematical Models

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  1. Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 Time (sec) Position (m) 0 10 1 12 2 14 3 16 4 20 5 26 Where is the runner: at 2 sec? at 3 sec? at 4½ sec? At what time was the runner at the 15m mark? What is his speed at 2 sec? 1 2 3 4 5 6 time (sec)

  2. Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 Time (sec) Position (m) 0 10 1 12 2 14 3 16 4 20 5 26 Where is the runner: at 2 sec? 14m at 3 sec? 16m at 4½ sec? 23m At what time was the runner at the 15m mark? 2.5 sec What is his speed at 2 sec? Velocity (speed) = the slope of the line on a distance vs time graph. 1 2 3 4 5 6 time (sec)

  3. Ex2) Describe each person’s motion. (What is their speed?) 1. 2. 3. 10 8 dist (m) 6 4 2 1 2 3 4 5 time (sec)

  4. Ex3) Write an equation that describes the motion of the airplane graphed, then use it to find its position at t = 2.5 sec. 200 160 dist (m) 120 80 40 1 2 3 4 5 time (sec)

  5. Ex4) a car starts 200m west of town square and moves at a constant vel of 15m/s east. A) Write an eqn to represent motion B) Where is the car after 2 min? When will the car reach town square? v = 15m/s Town

  6. Ex5) How fast is the plane going at 1.5 sec dist (m) time (sec)

  7. Ex5) How fast is the plane going at 1.5 sec dist (m) r time (sec) velocity = slope = = 6 m/s If the line curves, find the slope of the tangent line. Ch5 HW#1 1 – 4 rise = 6m run = 1sec

  8. Ch5 HW#1 1 – 5 1) How far is: - object A at t=3: t=8: -object B at t=4: t=8: -object C at t=5: t=10: 2) What is the velocity - of A - of B - of C

  9. Ch5 HW#1 1 – 5 1) How far is: - object A at t=3: 3m t=8: 8m -object B at t=4: 6m t=8: 4m -object C at t=5: 7m t=10: 7m 2) What is the velocity - of A - of B - of C

  10. Ch5 HW#1 1 – 5 1) How far is: - object A at t=3: 3m t=8: 8m -object B at t=4: 6m t=8: 4m -object C at t=5: 7m t=10: 7m 2) What is the velocity - of A 1m/s - of B -½m/s - of C 0m/s Object D starts at 0m and moves with constant velocity of 2m/s. What is its position after 4 sec?

  11. Ch5 HW#1 1 – 5 1) How far is: - object A at t=3: 3m t=8: 8m -object B at t=4: 6m t=8: 4m -object C at t=5: 7m t=10: 7m 2) What is the velocity - of A 1m/s - of B -½m/s - of C 0m/s Object D starts at 0m and moves with constant velocity of 2m/s. What is its position after 4 sec? df = di + vt 0m + (2m/s)(4s) = 8m 4. Object E starts 10m away and moves at –2m/s. When will it reach the start line?

  12. Ch5 HW#1 1 – 5 1) How far is: - object A at t=3: 3m t=8: 8m -object B at t=4: 6m t=8: 4m -object C at t=5: 7m t=10: 7m 2) What is the velocity - of A 1m/s - of B -½m/s - of C 0m/s Object D starts at 0m and moves with constant velocity of 2m/s. What is its position after 4 sec? df = di + vt 0m + (2m/s)(4s) = 8m 4. Object E starts 10m away and moves at –2m/s. When will it reach the start line? df = di + vt 0m = 10m + (–2m/s)(t) t = 5sec

  13. Ch5.2 Velocity Graphs Ex1) Explain the velocity of B: Ex2) Explain the velocity of A: Ex3) If B travels at 5m/s for 6 secs, how far did it travel? Ex4) Find the displacement of C after 4 sec. Ex5) Find the displacement of A after 3 sec. Ch5 HW#2 5 – 8 10 9 8 7 vel 6 (m/s) 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 time (sec)

  14. Lab5.1 – Ave vs Inst Acceleration - due tomorrow - Ch5 HW#2 due at beginning of period

  15. Ch5 HW#2 5 – 8 5) vel A at t=2 v=___ m/s vel B at t=4 v=___ m/s vel C at t=6 v=___ m/s 6) Displacement of A at t=5 sec: 7) Displacement of B at t=8 sec: 8) Displacement of C at t=10 sec:

  16. Ch5 HW#2 5 – 8 5) vel A at t=2 v=_4_ m/s vel B at t=4 v=_3_ m/s vel C at t=6 v=_2_ m/s 6) Displacement of A at t=5 sec: 7) Displacement of B at t=8 sec: 8) Displacement of C at t=10 sec:

  17. Ch5 HW#2 5 – 8 5) vel A at t=2 v=_4_ m/s vel B at t=4 v=_3_ m/s vel C at t=6 v=_2_ m/s 6) Displacement of A at t=5 sec: d = (area) = l.w = (5sec)(4m/s) = 20m 7) Displacement of B at t=8 sec: 8) Displacement of C at t=10 sec:

  18. Ch5 HW#2 5 – 8 5) vel A at t=2 v=_4_ m/s vel B at t=4 v=_3_ m/s vel C at t=6 v=_2_ m/s 6) Displacement of A at t=5 sec: d = (area) = l.w = (5sec)(4m/s) = 20m 7) Displacement of B at t=8 sec: d = (area) = ½b.h = ½(8sec)(8m/s) = 32m 8) Displacement of C at t=10 sec:

  19. Ch5 HW#2 5 – 8 5) vel A at t=2 v=_4_ m/s vel B at t=4 v=_3_ m/s vel C at t=6 v=_2_ m/s 6) Displacement of A at t=5 sec: d = (area) = l.w = (5sec)(4m/s) = 20m 7) Displacement of B at t=8 sec: d = (area) = ½b.h = ½(8sec)(8m/s) = 32m 8) Displacement of C at t=10 sec: d = (area) – (area) = ½b.h – ½b.h = ½(8sec)(8m/s) – ½(2sec)(2m/s) = 32m – 2m = 30m

  20. Ch 5.2 - Acceleration Units: HW#9) A race car’s velocity increases from 4.0m/s to 36m/s in 4 sec. What is it average acceleration?

  21. Ex9) Describe each motion: A: B: Find the acceleration of each : A: B: Find the distance traveled for each : A: B:

  22. Ex2) Describe the sprinter’s velocity : Describe the accl : What is the instantaneous accl a. t= 1 sec? b. t = 5 sec? What is the average accl from t = 0 – 10 sec? Ch5 HW#3 9 – 14 10 9 8 7 vel 6 (m/s) 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 time (sec)

  23. Ch5 HW#3 9 – 14 9. In class 10. A race car slows from 36 m/s to 15 m/s over 35sec. What was its average acceleration? vi = 36 m/s vf = 15 m/s t = 35 sec a = ? 11) A car is coasting backward downhill at a speed of 3 m/s when the driver gets the engine started. After 2.5 sec, the car is moving uphill at 4.5 m/s. If uphill is (+) what is the car’s average acceleration? vi = –3 m/s vf = +4.5 m/s t = 2.5 sec a = ?

  24. Ch5 HW#3 9 – 14 9. In class 10. A race car slows from 36 m/s to 15 m/s over 35sec. What was its average acceleration? vi = 36 m/s vf = 15 m/s t = 35 sec a = ? 11) A car is coasting backward downhill at a speed of 3 m/s when the driver gets the engine started. After 2.5 sec, the car is moving uphill at 4.5 m/s. If uphill is (+) what is the car’s average acceleration? vi = –3 m/s vf = +4.5 m/s t = 2.5 sec a = ?

  25. Ch5 HW#3 9 – 14 9. In class 10. A race car slows from 36 m/s to 15 m/s over 35sec. What was its average acceleration? vi = 36 m/s vf = 15 m/s t = 35 sec a = ? 11) A car is coasting backward downhill at a speed of 3 m/s when the driver gets the engine started. After 2.5 sec, the car is moving uphill at 4.5 m/s. If uphill is (+) what is the car’s average acceleration? vi = –3 m/s vf = +4.5 m/s t = 2.5 sec a = ?

  26. 12) A bus is moving at 25 m/s when the driver steps on the breaks and brings the bus to a stop in 3.0 sec. a. What is the average acceleration? b. If it took twice as long to stop, what would be the accl? vi = 25 m/s vf = 0 m/s t = 3.0 sec a = ? 13)Look at the graph: a. Where is the speed constant? b. Over what intervals is accl positive? c. Over what intervals is accl negative? 14) Find average acceleration at: a. 0 to 5 sec b. 15 to 20 sec c. 0 to 40 sec 10 8 6 4 2 vel (m/s) 10 20 30 40 50 time (sec)

  27. 12) A bus is moving at 25 m/s when the driver steps on the breaks and brings the bus to a stop in 3.0 sec. a. What is the average acceleration? b. If it took twice as long to stop, what would be the accl? vi = 25 m/s vf = 0 m/s t = 3.0 sec a = ? 13)Look at the graph: a. Where is the speed constant? b. Over what intervals is accl positive? c. Over what intervals is accl negative? 14) Find average acceleration at: a. 0 to 5 sec b. 15 to 20 sec c. 0 to 40 sec 10 8 6 4 2 vel (m/s) 10 20 30 40 50 time (sec)

  28. 12) A bus is moving at 25 m/s when the driver steps on the breaks and brings the bus to a stop in 3.0 sec. a. What is the average acceleration? b. If it took twice as long to stop, what would be the accl? vi = 25 m/s vf = 0 m/s t = 3.0 sec a = ? 13)Look at the graph: a. Where is the speed constant? 5-15sec b. Over what intervals is accl positive? 0-5 c. Over what intervals is accl negative? 15-20s, 25-40s 14) Find average acceleration at: a. 0 to 5 sec b. 15 to 20 sec c. 0 to 40 sec 10 8 6 4 2 vel (m/s) 10 20 30 40 50 time (sec)

  29. 12) A bus is moving at 25 m/s when the driver steps on the breaks and brings the bus to a stop in 3.0 sec. a. What is the average acceleration? b. If it took twice as long to stop, what would be the accl? vi = 25 m/s vf = 0 m/s t = 3.0 sec a = ? 13)Look at the graph: a. Where is the speed constant? 5-15sec b. Over what intervals is accl positive? 0-5 c. Over what intervals is accl negative? 15-20s, 25-40s 14) Find average acceleration at: a. 0 to 5 sec (slope) a = 2m/s2 b. 15 to 20 sec (slope) a = -6/5 m/s2 c. 0 to 40 sec 10 8 6 4 2 vel (m/s) 10 20 30 40 50 time (sec)

  30. Ch5.3 – Motion Equations 1.df= di + v∙t(constant velocity, a = 0)

  31. Ch5.3 – Motion Equations 1.df= di + v∙t(constant velocity, a = 0) 2.vf = vi + a∙t (accelerating, independent of distance)

  32. Ch5.3 – Motion Equations 1.df= di + v∙t (constant velocity, a = 0) 2.vf = vi + a∙t (accelerating, independent of distance) 3. d = ½(vi +vf ).t (accelerating, independent of the accl)

  33. Ch5.3 – Motion Equations 1. df= di + v∙t (constant velocity, a = 0) 2. vf = vi + a∙t (accelerating, independent of distance) 3. d = ½(vi +vf ).t (accelerating, independent of the accl) 4. df = vit + ½a.t2 (accelerating, independent of vf)

  34. Ch5.3 – Motion Equations 1. df= di + v∙t (constant velocity, a = 0) 2. vf = vi + a∙t (accelerating, independent of distance) 3. d = ½(vi +vf ).t (accelerating, independent of the accl) 4. df = vit + ½a.t2 (accelerating, independent of vf) 5. vf2 = vi2 + 2.a.d (accelerating, independent of time) To solve motion problems, must be given 3 variables then solve for a 4th. Ex1) What is the final velocity of a car that accelerates from a stop at 3.5m/s2 for 4 sec?

  35. Ch5.3 – Motion Equations 1. df= di + v∙t (constant velocity, a = 0) 2. vf = vi + a∙t (accelerating, independent of distance) 3. d = ½(vi +vf ).t (accelerating, independent of the accl) 4. df = vit + ½a.t2 (accelerating, independent of vf) 5. vf2 = vi2 + 2.a.d (accelerating, independent of time) Ex2) A car is going 20m/s when it uniformly accelerates to 25m/s in 2 sec. What distance does it travel?

  36. Ch5.3 – Motion Equations 1. df= di + v∙t (constant velocity, a = 0) 2. vf = vi + a∙t (accelerating, independent of distance) 3. d = ½(vi +vf ).t (accelerating, independent of the accl) 4. df = vit + ½a.t2 (accelerating, independent of vf) 5. vf2 = vi2 + 2.a.d (accelerating, independent of time) HW#15) A golf ball rolls up hill. a. If it starts at 2.0m/s and shows at a constant . 50 m/s2 what is its velocity after 2.0 sec? b. If the ball continues for 6 sec, what is its velocity?

  37. Ch5.3 – Motion Equations 1. df= di + v∙t (constant velocity, a = 0) 2. vf = vi + a∙t (accelerating, independent of distance) 3. d = ½(vi +vf ).t (accelerating, independent of the accl) 4. df = vit + ½a.t2 (accelerating, independent of vf) 5. vf2 = vi2 + 2.a.d (accelerating, independent of time) Ex2) A car starts at rest and speeds up at 3.5 m/s2 . How far will it have gone when it is going 25 m/s?

  38. Ch5.3 – Motion Equations 1. df= di + v∙t (constant velocity, a = 0) 2. vf = vi + a∙t (accelerating, independent of distance) 3. d = ½(vi +vf ).t (accelerating, independent of the accl) 4. df = vit + ½a.t2 (accelerating, independent of vf) 5. vf2 = vi2 + 2.a.d (accelerating, independent of time) HW#19) A race car traveling at 44m/s slows at a constant rate to a velocity of 22m/s over 11sec. How far does it move over that time? Ch5 HW#4 15 – 22

  39. Lab5.2 – Motion on an Incline Plane - due tomorrow - Ch5 HW#4 due at beginning of period

  40. Ch5 HW#4 15 – 22 15. In class 16. A bus travelling at 30 m/s, speeds up at a constant rate of 3.5 m/s2. What is its velocity 6.8 sec later? vi = vf = t = a = 17. If a car accelerates from rest at a constant 5.5 m/s2, how long will it need to reach 28 m/s?

  41. Ch5 HW#4 15 – 22 15. In class 16. A bus travelling at 30 m/s, speeds up at a constant rate of 3.5 m/s2. What is its velocity 6.8 sec later? vi = 30 m/svf = vi + a ∙ t vf = ? t = 6.8 sec a = 3.5 m/s2 17. If a car accelerates from rest at a constant 5.5 m/s2, how long will it need to reach 28 m/s? vi = 0 m/svf = vi + a ∙ t vf = 28 m/s a = 5.5 m/s2 t = ? or d = ? vf2 = vi2 + 2.a.d

  42. Ch5 HW#4 15 – 22 18. A car slows from 22 m/s to 3 m/s at a constant rate of 2.1 m/s2 . How much time did this take? vi = vf = a = t = 20. A car accelerates at a constant rate from 15 m/s to 25 m/s while it travels 125m. How much time does this take? vi = vf = d = t =

  43. Ch5 HW#4 15 – 22 18. A car slows from 22 m/s to 3 m/s at a constant rate of 2.1 m/s2 . How much time did this take? vi = 22 m/svf = vi + a ∙ t vf = 3 m/s a = –2.1 m/s2 t = ? 20. A car accelerates at a constant rate from 15 m/s to 25 m/s while it travels 125m. How much time does this take? vi = 15 m/sd = ½(vi +vf ).t vf = 25 m/s d = 125 m t = ?

  44. A bike rider accelerates constantly to a velocity of 7.5 m/s during 4.5 sec. The bikes displacement during that accl is 19m. What is the initial velocity of the bike? vi = ? vf = t = d = 22. An airplane starts from rest and accelerates at a constant rate of 3.00 m/s2 for 30 seconds a. How far did it move? b. How fast was it going? vi = a = t = d = ? vf = ?

  45. A bike rider accelerates constantly to a velocity of 7.5 m/s during 4.5 sec. The bikes displacement during that accl is 19m. What is the initial velocity of the bike? vi = ? d = ½(vi +vf ).t vf = 7.5 m/s t = 4.5s d = 19m 22. An airplane starts from rest and accelerates at a constant rate of 3.00 m/s2 for 30 seconds a. How far did it move? b. How fast was it going? vi = 0 m/sdf = vit + ½a.t2 vf = vi + a ∙ t a = 3.00 m/s2 t = 30 sec d = ? vf = ?

  46. Ch 5.4 – Free Fall Neglecting air resistance, all objects fall at the same rate. (Same acceleration.) Doesn’t matter if objects are thrown upward, thrown downward, across, etc, their acceleration is toward the center of the earth. a = g = 9.8 m/s2 (You are authorized to use 10 m/s2.) Ex1) Demon drop falls freely for 1.5 sec, after starting from rest. a. What is its velocity at 1.5 sec? b. How far does it fall?

  47. Ex2) A bullet is fired straight up at 250 m/s. a. How high does it go? b. How much time is it in the air? c. How fast is it going when it comes back down?

  48. Graphs Ex3) A ball is dropped from a hovering helicopter. Graph its velocity and displacement for the 1st 5 sec. vf = vi + a ∙ t d = ½(vi +vf ).t t = 1 t = 2 t = 3 t = 4 t = 5 80 70 60 50 40 30 20 10 dist (m) 50 40 30 20 10 vel (m/s) 1 2 3 4 5 time (sec) 1 2 3 4 5 time (sec) Ch5 HW#5 23-25 1 2 3 4 5 time (sec)

  49. Lab5.3 – Measuring ‘g’ - due tomorrow - Ch5 HW#5 due at beginning of period - Ch5 HW#6 due tomorrow

  50. Ch5 HW#5 23-25 23) A brick is dropped, a. What is its velocity after 4 sec? b. How far does it fall?vf = vi + a ∙ t df = vit + ½a.t2 vi = 0 m/s a = 9.8 m/s2 t = 4 sec vf = ? d = ? 24) Tennis ball thrown straight up with an initial speed of 22.5 m/s. a. How high does it rise? vf = 0 d = ? ttotal = ? b. How long till caught again? a = –9.8 m/s2 vi = 22.5 m/s

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