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Rules of This Activity #1 All students must perform all of the work of the activity in their science journals . #2 When you are sure of your answer, call the teacher/student teacher over to confirm. #3 If it is confirmed that you are correct, become a student teacher and help someone in class who does not understand the method used to solve the problem. #4 If you do not understand how to begin to answer the question, raise your hand!
Stating the Factoids Batfink 50 kg Karate 150 kg Hugo A-go-go 75 kg Any other character mentioned will be considered 65 kg  Vssr = 340 m/s Batallac 1500 kg Mobile City Destroyer 2500 kg Any other vehicle will be considered 1250 kg
Calculate Gus’ Acceleration Greasy Gus jumps in his un-started car.  In 1.76s, he and the car are speeding through the open door 10 meters away.
What is Greasy Gus’ acceleration? a = vf–vo/tf-to     Known			                    Want m = 1,250 kg 	                                a  vo = 0 m/s t =  1.76s   vf = 5.68 m/s a = (5.68 – 0 m/s )/(1.76 s)  a = 3.23 m/s2
Working the Problem Graph diagrams of displacement versus time, velocity verus time, and acceleration versus time for your calculations from the previous problem.
Motion Graphing     x			   v		          a                                                                                        t
Working the Problem What was the force (N) needed to accelerate this car and occupant at that rate?
What was the force (N) needed to accelerate this car and occupant at that rate?     Known			        Want mc = 1,250 kg 	                    F  mo = 65 kg  a = 3.23 m/s2 F = ma F = (1,315 kg )(3.23 m/s2)  F = 4247.45 (kg)(m/s2)
Working the Problem Batfink sends a super sonic sonar  		                         out to find Greasy Gus’ hideout.
Working the Problem How far away (Xf)is Greasy Gus’ Hideout if it took the super sonic radar “BEEP” nine seconds to return to Batfink’s ear?  Xf
Xfof Gus’ Hideout Known			        Want xo = o m 	                               Xf vo = 340 m/s  a = 0 m/s2 t =  4.5 s Xf = Xo+ vot + ½ at2 Xf = 0 + (340 m/s)(4.5 s)+ 0 Xf = 1530 m
Stating the Problem The Batallac continuously accelerates at 2.25 m/s2, how long will it take it      to drive the same     distance as the super    sonic sonar radar beep   just traveled?
Time to Gus’ Hideout  via Battalac Known			        Want xo = o m 	                               t Vo = 0 m.s a = 2.25 m/s2 Xf = 1530 m 1530 m = 0 m + 0 + ½ (2.25 m/s2 )(t)2 1530 m = 0 + (1.13 m/s2)(t) 2 t = 36.79 s
Stating the Problem Since the Batallac has to stop before crashing into the hideout, you will need to calculate the vf of the Battalac.
Working the Problem Known			        Want xo = o m 	                              Vf Vo = 0 m.s a = 2.25 m/s2 Xf = 1530 m                                 Vf2 = Vo2 +2a∆x                               Vf2 = 0 + (2)(2.25 m/s2 )(1530 m) Vf= 82.98 m/s
Stating the Problem The Batallac is equipped with Plutonium encrusted brake pads capable of decelerating the vehicle at 25 m/s2.  How long will it take the Battalac to come to a complete stop once the brakes are fully applied?
Working the Problem Known			                 Want xo = o m 	                              t  Vo = 82.98 m/s Vf = 0 m.s a = -25 m/s2 Vf = Vo + at        0 = 82.98 m/s + (- 25 m/s2)(t)                       t =  3.32 s
Stating the Problem What is the momentum of the Battalac just before applying the brakes?
Working the Problem? P = mv     Known			                            Want m = 1,700 kg 	                                P (momentum)  v = 82.98 m/s P = mv P = (1,700 kg )(82.98 m/s)  P = 141,066 (kg)( m/s)
Stating the Problem Karate and Batfink are trapped in tires.  They are rolling at 3 m/s when they fall off the edge of the cliff 135m in height.
Working the Problem How long will Batfink and Karate be in the air?
Working the Problem. ∆y = ½ gt2     Known			                            Want g = -9.8 m/s2 	                                          t  ∆y = 135m ∆y = ½ gt2 135m = ½ (-9.8m/s2)(t)2 t = 5.25 s
Working the Problem How far away from the cliff base will Batfink and Karate land?
Working the Problem Xf = xo + Vot + ½ at2     Known			                            Want a = 0.0 m/s2Xf Xo= 0m Vo = 3.0 m/s t =  5.25 s Xf = xo + Vot + ½ at2 Xf = 0 + (3 m/s)(5.25s) + 0 Xf = 15.75m
Stating the Problem What will be the Vyf of Batfink and Karate just before impact with the ground?
Vyf of Batfink and Karate Vf2 = Vo2 + 2a ∆y     Known			                            Want a = 9.8 m/s2Vyf Xo= 0m Vo = 0.0 m/s t =  5.25 s ∆y = 135m Vf2 = 0 + 2(9.8 m/s2)(135m) Vf = 51.44m/s
Working the Problem What is the magnitude of Batfink and Karate’s  velocity just before they would make contact with the ground?
The magnitude of Batfink and Karate’s  velocity. a2 + b2 = c2 51.442 + 3.02 = 51.52 m/s                      y 3.0 m/s 51.44 m/s51.52 m/s                    				x
State the Problem Draw 5 vector diagrams that will illustrate the fall of Batfink or Karate off of the cliff. Vector Diagram TEmplate Horizontal Velocity Vertical Velocity Force Acceleration Momentum 0s  Not rolling on top of cliff 1s  Rolling on top of cliff	 4s  Falling 6s  Falling
Draw Vector Diagrams HORIZONTAL vELOCITY VERTICAL VELOCITY y			          	                                          x
Draw Vector Diagrams force acceleration  y                                             x y                                               x
Draw Vector Diagrams momentum What does the area under this graph represent? Y                                         x ∆v                                            t
State the Problem Batfink and Karate are waiting for Greasy Gus at the turn of a road.   The Batallaccovered the 10km in 333.33s.  What was the Batallac’s averagespeed?  Greasy Gus took 15s longer.  What was his average speed?
Average Speed V = d/t     Known			                            Want d = 10,000m 	                                         V Xo= 0m Vo = 0.0 m/s tb =  333.33s tg =  348.33s Vb = d/t		           Vg = d/t	         V = 10,000m/333.33s        V = 10,000m/348.33s         V =  30.00 m/s                     V =  28.71 m/s
Stating the Problem If the two cars started at the same time and each car  accelerated uniformly at 3 m/s2, how far from the turn in the road was Gus when the Battalac had already reached the turn?
How far from the turn in the road was Gus? d = (v)(t)     Known			                            Want Vb = 30.00 m/s	                                         d Vg = 28.71 m/s tb =  333.33s                                     d = (v)(t)                       d = (28.71 m/s)(333.33s)                                  dg = 9569.90                          d = 10,000 – 9569.90                                    d = 430.1m
Working the Problem Driving recklessly, Gus ends up driving his car up a tree.  If he was driving 28.71 m/s, how are up the tree will his car end up?  Assume the tree is at a 90* angle.
How far up the tree will Gus’ car end up?                      t = v/a                                 Known			                            Want Vb = 28.71 m/s	                                         y g = -9.8m/s2 tg =  2.93s yf = yo + vot + ½ gt2 yf = 0 + (28.71m/s)(2.93s) + ½ (-9.8m/s2)(2.93s)2  yf = 42.08m
Stating the Problem If Gus’ car falls from the tree, how much momentum will it have when it strikes the ground?
Gus’ Car’s Momentum P = mv Known			               Want mg = 65.00 kg	                                         P mc =  1250 kg Vg = 28.71 m/s                             P = mv   P = (65.00 kg + 1250 kg) (28.71 m/s) P = 37,753.65 kg m/s
This Semester Exam Cannot Harm me.  My Wings are like a shield of steel!

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Ppt batfink semester 1 review

  • 1.
  • 2. Rules of This Activity #1 All students must perform all of the work of the activity in their science journals . #2 When you are sure of your answer, call the teacher/student teacher over to confirm. #3 If it is confirmed that you are correct, become a student teacher and help someone in class who does not understand the method used to solve the problem. #4 If you do not understand how to begin to answer the question, raise your hand!
  • 3. Stating the Factoids Batfink 50 kg Karate 150 kg Hugo A-go-go 75 kg Any other character mentioned will be considered 65 kg Vssr = 340 m/s Batallac 1500 kg Mobile City Destroyer 2500 kg Any other vehicle will be considered 1250 kg
  • 4. Calculate Gus’ Acceleration Greasy Gus jumps in his un-started car. In 1.76s, he and the car are speeding through the open door 10 meters away.
  • 5. What is Greasy Gus’ acceleration? a = vf–vo/tf-to Known Want m = 1,250 kg a vo = 0 m/s t = 1.76s vf = 5.68 m/s a = (5.68 – 0 m/s )/(1.76 s) a = 3.23 m/s2
  • 6. Working the Problem Graph diagrams of displacement versus time, velocity verus time, and acceleration versus time for your calculations from the previous problem.
  • 7. Motion Graphing x v a t
  • 8. Working the Problem What was the force (N) needed to accelerate this car and occupant at that rate?
  • 9. What was the force (N) needed to accelerate this car and occupant at that rate? Known Want mc = 1,250 kg F mo = 65 kg a = 3.23 m/s2 F = ma F = (1,315 kg )(3.23 m/s2) F = 4247.45 (kg)(m/s2)
  • 10. Working the Problem Batfink sends a super sonic sonar out to find Greasy Gus’ hideout.
  • 11. Working the Problem How far away (Xf)is Greasy Gus’ Hideout if it took the super sonic radar “BEEP” nine seconds to return to Batfink’s ear? Xf
  • 12. Xfof Gus’ Hideout Known Want xo = o m Xf vo = 340 m/s a = 0 m/s2 t = 4.5 s Xf = Xo+ vot + ½ at2 Xf = 0 + (340 m/s)(4.5 s)+ 0 Xf = 1530 m
  • 13. Stating the Problem The Batallac continuously accelerates at 2.25 m/s2, how long will it take it to drive the same distance as the super sonic sonar radar beep just traveled?
  • 14. Time to Gus’ Hideout via Battalac Known Want xo = o m t Vo = 0 m.s a = 2.25 m/s2 Xf = 1530 m 1530 m = 0 m + 0 + ½ (2.25 m/s2 )(t)2 1530 m = 0 + (1.13 m/s2)(t) 2 t = 36.79 s
  • 15. Stating the Problem Since the Batallac has to stop before crashing into the hideout, you will need to calculate the vf of the Battalac.
  • 16. Working the Problem Known Want xo = o m Vf Vo = 0 m.s a = 2.25 m/s2 Xf = 1530 m Vf2 = Vo2 +2a∆x Vf2 = 0 + (2)(2.25 m/s2 )(1530 m) Vf= 82.98 m/s
  • 17. Stating the Problem The Batallac is equipped with Plutonium encrusted brake pads capable of decelerating the vehicle at 25 m/s2. How long will it take the Battalac to come to a complete stop once the brakes are fully applied?
  • 18. Working the Problem Known Want xo = o m t Vo = 82.98 m/s Vf = 0 m.s a = -25 m/s2 Vf = Vo + at 0 = 82.98 m/s + (- 25 m/s2)(t) t = 3.32 s
  • 19. Stating the Problem What is the momentum of the Battalac just before applying the brakes?
  • 20. Working the Problem? P = mv Known Want m = 1,700 kg P (momentum) v = 82.98 m/s P = mv P = (1,700 kg )(82.98 m/s) P = 141,066 (kg)( m/s)
  • 21. Stating the Problem Karate and Batfink are trapped in tires. They are rolling at 3 m/s when they fall off the edge of the cliff 135m in height.
  • 22. Working the Problem How long will Batfink and Karate be in the air?
  • 23. Working the Problem. ∆y = ½ gt2 Known Want g = -9.8 m/s2 t ∆y = 135m ∆y = ½ gt2 135m = ½ (-9.8m/s2)(t)2 t = 5.25 s
  • 24. Working the Problem How far away from the cliff base will Batfink and Karate land?
  • 25. Working the Problem Xf = xo + Vot + ½ at2 Known Want a = 0.0 m/s2Xf Xo= 0m Vo = 3.0 m/s t = 5.25 s Xf = xo + Vot + ½ at2 Xf = 0 + (3 m/s)(5.25s) + 0 Xf = 15.75m
  • 26. Stating the Problem What will be the Vyf of Batfink and Karate just before impact with the ground?
  • 27. Vyf of Batfink and Karate Vf2 = Vo2 + 2a ∆y Known Want a = 9.8 m/s2Vyf Xo= 0m Vo = 0.0 m/s t = 5.25 s ∆y = 135m Vf2 = 0 + 2(9.8 m/s2)(135m) Vf = 51.44m/s
  • 28. Working the Problem What is the magnitude of Batfink and Karate’s velocity just before they would make contact with the ground?
  • 29. The magnitude of Batfink and Karate’s velocity. a2 + b2 = c2 51.442 + 3.02 = 51.52 m/s y 3.0 m/s 51.44 m/s51.52 m/s x
  • 30. State the Problem Draw 5 vector diagrams that will illustrate the fall of Batfink or Karate off of the cliff. Vector Diagram TEmplate Horizontal Velocity Vertical Velocity Force Acceleration Momentum 0s Not rolling on top of cliff 1s Rolling on top of cliff 4s Falling 6s Falling
  • 31. Draw Vector Diagrams HORIZONTAL vELOCITY VERTICAL VELOCITY y x
  • 32. Draw Vector Diagrams force acceleration y x y x
  • 33. Draw Vector Diagrams momentum What does the area under this graph represent? Y x ∆v t
  • 34. State the Problem Batfink and Karate are waiting for Greasy Gus at the turn of a road. The Batallaccovered the 10km in 333.33s. What was the Batallac’s averagespeed? Greasy Gus took 15s longer. What was his average speed?
  • 35. Average Speed V = d/t Known Want d = 10,000m V Xo= 0m Vo = 0.0 m/s tb = 333.33s tg = 348.33s Vb = d/t Vg = d/t V = 10,000m/333.33s V = 10,000m/348.33s V = 30.00 m/s V = 28.71 m/s
  • 36. Stating the Problem If the two cars started at the same time and each car accelerated uniformly at 3 m/s2, how far from the turn in the road was Gus when the Battalac had already reached the turn?
  • 37. How far from the turn in the road was Gus? d = (v)(t) Known Want Vb = 30.00 m/s d Vg = 28.71 m/s tb = 333.33s d = (v)(t) d = (28.71 m/s)(333.33s) dg = 9569.90 d = 10,000 – 9569.90 d = 430.1m
  • 38. Working the Problem Driving recklessly, Gus ends up driving his car up a tree. If he was driving 28.71 m/s, how are up the tree will his car end up? Assume the tree is at a 90* angle.
  • 39. How far up the tree will Gus’ car end up? t = v/a Known Want Vb = 28.71 m/s y g = -9.8m/s2 tg = 2.93s yf = yo + vot + ½ gt2 yf = 0 + (28.71m/s)(2.93s) + ½ (-9.8m/s2)(2.93s)2 yf = 42.08m
  • 40. Stating the Problem If Gus’ car falls from the tree, how much momentum will it have when it strikes the ground?
  • 41. Gus’ Car’s Momentum P = mv Known Want mg = 65.00 kg P mc = 1250 kg Vg = 28.71 m/s P = mv P = (65.00 kg + 1250 kg) (28.71 m/s) P = 37,753.65 kg m/s
  • 42. This Semester Exam Cannot Harm me. My Wings are like a shield of steel!