x = x 0 + vt x = 6000 m + (0.7 m/s)(25 min*60 s/min) = 7050 m I would be between the 7000m mark and the 8000m mark, closer to the 7000m mark.

Transcription

1 PUM Physics II - Kinematics Lesson 6 Solutions Page 1 of Observe and Represent a) The cars were only next to each other at the initial clock reading. This is because Car 2 was moving twice as fast as Car 1. The spacing between dots for Car 2 is twice as wide as for Car 1 meaning that in one second Car 2 went twice as far on the floor compared to Car 1. b) Eugenia circled the dots because they were next to each other on the diagram. However, beyond the first pair of dots the cars were at subsequent dot locations as different times. I would help Eugenia understand my point of view by asking her how often the beanbags were being dropped for each car. Since they were being dropped every second for both cars this must mean that the two cars were moving at different constant velocities, meaning past the initial point the two cars could never be next to each other at the same clock reading. c) Not enough information is provided to know which directions the cars were moving. The diagram doesn t indicate where the cars started. d) Car 1 Car 2 e) The two equations would have the same format with different subscripts perhaps. However, Car 2 was traveling at twice the velocity as Car 1. Therefore the equation for Car 2 would look like: x 2 (t) = 2v 1 t. 6.2 Represent and Reason v = 0.7 m/s Path marked ever 1000 m x 0 =6000 m x = x 0 + vt x = 6000 m + (0.7 m/s)(25 min*60 s/min) = 7050 m I would be between the 7000m mark and the 8000m mark, closer to the 7000m mark.

2 PUM Physics II - Kinematics Lesson 6 Solutions Page 2 of Represent and Reason a) Position vs. Time Graph b) Find the steepness of the position vs. time graph: x/ t = (-10 m 30 m)/(10 s 0 s) = -4.0 m/s Assuming Darshan stays at this constant velocity. x = x 0 + vt (-85.0 m) = 30.0 m + (-4.0 m/s)t t = ( m)/(-4.0 m/s) = 29 s c) Dot Diagram: d) III) m; the problem does not specify but how far in this case means the displacement. 6.4 Equation Jeopardy a) -62 mi/hr Mile Marker x 0 Mile Marker 114

3 PUM Physics II - Kinematics Lesson 6 Solutions Page 3 of 6 b) The Mystery Machine is traveling towards the west on route 44 at a constant velocity of - 62 miles/hour. After traveling 0.35 hours they find themselves at mile marker 114. At which mile marker did they begin their journey? c) Dot Diagram: d) Position vs. Time Graph e) 114mi = (!62 mi hr )(0.35hr) + x 0 x 0 = 114 mi + (62 mi/hr)(0.35 hr) = 136 mi 6.5 Represent and Reason Average Speed: Going there: (200 km)/(90 km/hr) = 2.2 hr Going back: (200 km)/(50 km/hr) = 4.0 hr Path Length / Total Time = (400 km)/(2.2hr + 1.0hr + 4.0hr) = 56 km/hr Average Velocity: Displacement / Total Time = (0 km)/(7.2 hr) = 0 km/hr

4 PUM Physics II - Kinematics Lesson 6 Solutions Page 4 of 6 a) 1hour break b) All unknown quantities were solved for in calculating the average speed. Everything else was given. 6.6 Practice a) Average Speed: First 30 min: (55 mi/hr)(0.5 hr) = 27.5mi Next 30 min: (75 mi/hr)(0.5 hr) = 37.5mi (65 mi)/( 1 hr) = 65 mi/hr b) Average Speed: First 30 mi: (30 mi)/(55 mi/hr) = 0.55hr Next 30 mi: (30 mi)/(75 mi/hr) = 0.40hr (60 mi)/(0.9 hr) = 63 mi/hr 6.7 Equation Jeopardy a) Equation: x = (!62.0 mi )(0.35hr) mi hr -62 mi/hr Mile Marker 4 Mile Marker x b) The Mystery Machine turns onto another road and travels for 0.35 hr to the west at mi/hr. If they started at the 4 mi marker where will they end up at the end of their trip?

5 PUM Physics II - Kinematics Lesson 6 Solutions Page 5 of 6 c) Dot Diagram: d) Position vs. Time Graph e) x = (!62.0 mi )(0.35hr) mi = -18 mile marker hr f) Equation: 114mi = (!v mi )(1.2hr) + (!30.0 mi) hr v mi 114mi Scooby is in trouble! The Mystery Machine starts at the -30 mi marker. Scooby is at the 114 mi marker. If the Mystery Machine gets there in 1.2 hr how fast were they going down the highway? Did they break the speed limit? What assumptions did you make? Dot Diagram:

6 PUM Physics II - Kinematics Lesson 6 Solutions Page 6 of 6 Position vs. Time Graph: 6.8 Practice (very challenging!) a) We don t know what the time is so let s say that the car moved for t minutes (or hours units don t matter here) at 55mph and t minutes at 75mph taking a total of 2t minutes. Distance = Speed*Time Average Speed = Path Length / Total Time [(55 mi/hr)t + (75 mi/hr)t]/2t t(55 mi/hr + 75 mi/hr)/2t (55 mi/hr + 75 mi/hr)/2 (130 mi/hr)/2 = 65 mi/hr b) This time we don t know the distance so let s say the car moved d distance at 55mph then another d distance at 75mph for a total path length of 2d: Time = Distance/Speed Average Speed = Path Length / Total Time = 63 mi/hr