# 7.1 Represent and Reason a) The bike is moving at a constant velocity of 4 m/s towards the east

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1 PUM Physics II - Kinematics Lesson 7 Solutions Page 1 of Represent and Reason a) The bike is moving at a constant velocity of 4 m/s towards the east b) For the same motion, a position versus time graph would be a straight line at a diagonal. Where the slope would be 4 m/s. Position instead of velocity would be on the y axis. c) By finding the area between the line and the axis from the 10s to the 15s marks, we can estimate the bike s displacement. The area between the line and the axis will be a rectangle with sides v and t. The area of a rectangle is the product of its sides so v* t. This makes sense since v = x/ t therefore x = v* t. Also the units make sense [m/s]*[s] = m. d) Bikes diplacement from 0 to 20 seconds Area under line between t=0 s and t=20 s (4 m/s)(20 s) = 80 m e) The object s displacement can be found by finding the area between the velocity line and the x-axis during some time interval if the object is moving at a constant velocity. f) Yes, the mathematical model is consistent with our rule for finding the displacement graphically from a velocity vs time graph. However, we cannot from the velocity versus time graph however determine the position unless we assume x 0 = 0. Otherwise the rule and mathematical model are the same. 7.2 Represent and Reason a) The bike starts at a position of -4 m and travels at a constant -4 m/s towards -36 m, the person on the bike then stops and takes a break for 7 s, and turns around and then travels at a non-constant, faster speed than previous until the 18 m mark where the bike starts to slow down until it reaches +30 m. The person on the bike then stops for a brief 2 s break, and then turns around again and travels at a constant -5 m/s towards a position of 0 meters. The bike rider then takes another 4 second break, and pedals at a slower, constant -2 m/s until she reaches -16 m.

2 PUM Physics II - Kinematics Lesson 7 Solutions Page 2 of 7 b) Velocity vs. Clock Reading 7.3 Represent and Reason 6.0 Velocity vs. Time Velocity (m/s) time (s)

3 PUM Physics II - Kinematics Lesson 7 Solutions Page 3 of 7 a) The hiker walks south at a constant velocity of 2.5 m/s from 0 s to about 50 s, from 50 s to about 60 s the hiker takes a break before moving again to the south at 4.0 m/s from 60 s to 110 s b) The hiker moved for 15 s between the 10 s and 25 s clock reading at a velocity of 2.5 m/s. Using the graph we can find the area traced out by this rectangle: (2.5 m/s)(15 s) = 37.5 m c) To find how far the hiker moved from 40 to 70 seconds, we find the area of the two rectangles while he was moving and add these together. The first is from 40 s to 50 s: (2.5 m/s)(10 s) = 25 m The second is from 60s to 70s: (4.0 m/s)(10 s) = 40 m. The total displacement of the hiker is 65m meters. d) The average speed of the hiker is the Path Length / Trip Time. Because the hiker is hiking along a straight path, the path length is equal to the displacement. The total displacement for the hiker is the total area of both rectangles: 7.4 Evaluate First rectangle: (2.5m/s)(50s) = 125 m Second rectangle: (4.0m/s)(50s) = 200 m Total Displacement = 325 m Total Time = 110 s Average Speed = (325 m)/(110 s) = 3.38 m/s Stop and read the graphs! Notice that Bike A is a velocity vs time graph while Bike B is a position vs time graph. d) Bike B stopped twice during the trip. Since Bike B is a position vs time graph the two parts of the graph where the line is completely horizontal indicate that the bike has 0 velocity during these points. The bike s position stays constant over time during these intervals. g) The last part of the trip bike A was moving at constant speed in the negative direction. Since Bike A is a velocity vs time graph, during the last part of the trip the line is below the x-axis meaning that the bike was traveling in the negative direction. The speed is constant since the line is completely horizontal, meaning that the speed does not change with time during this interval.

4 PUM Physics II - Kinematics Lesson 7 Solutions Page 4 of 7 h) When we started observing Bike B it was moving at constant positive velocity. Initially the line has a positive slope which indicates a positive velocity on a position vs time graph. i) When we started observing bike A it was moving at increasing velocity in the positive direction, then it reached some constant velocity (positive) and continued moving for a while, then its velocity started decreasing and it some point it became zero. The it continued to increased in the negative direction until it reached some new velocity which it maintained for a while. On a velocity vs time graph constantly increasing velocity is indicated by a positive slope whereas constantly decreasing velocity is indicated by a negative slope, finally constant velocity is indicated by 0 slope. Given this the graph matches the description. j) When we started observing bike B it was moving at constant velocity in the positive direction, then it stopped for a while, then it started going back to the origin and then in the negative direction. Finally it stopped. On a position vs time graph constant velocity in the positive direction is indicated by a positive slope whereas constant velocity in the negative direction is indicated by a negative slope, finally zero velocity is indicated by 0 slope. Given this the graph matches the description. 7.5 Represent and Reason Homework a) Total Trip Time = 3 hr 20 min = 3.3hr First part of trip: (130 mi)/(65 mi/hr) = 2.0 hr Second part of trip: (3.3 hr 2.0 hr) = 1.3 hr (1.3 hr)(55 mi/hr) = 72 mi 65 mi/hr 55 mi/hr Home University of Delaware 130 mi 202mi Average Speed: Path Length / Time = 202 mi / 3.3 hr = 61 mi/hr Average Velocity: Assuming the car traveled in the same direction Path Length = Displacement so Average Velocity = Average Speed

5 PUM Physics II - Kinematics Lesson 7 Solutions Page 5 of 7 b) On this scale it is difficult to notice the slight change in slope from hrs c) Time (hr) 7.6 Evaluate a) The first two graphs (A and B) provide the same information. A is a position vs time graph while B is a velocity vs time graph. A says that the object stayed at the same positive position for 4 s while B says the object moved at a constant positive velocity for 4 s. A person might choose this wrong answer as correct because the two graphs look the same despite having different axis. b) The second two graphs (C and D) provide the same information. D is a position vs time graph while C is a velocity vs time graph. A says that the object stayed at the same negative position for 4 s while B says the object moved at a constant negative velocity for 4 s. A person might choose this wrong answer as correct because the two graphs look the same despite having different axis.

6 PUM Physics II - Kinematics Lesson 7 Solutions Page 6 of 7 c) Object A traveled 60 meters in 3 seconds from the location is was at the 0 clock reading. This would be true if A was a velocity vs. time graph, which is what a person choosing e) Object C was not moving during the experiment. This would be true if C was a position vs. time graph, which is what a person choosing h) Object D was moving in the negative direction at the speed of 20 m/s This would be true if D was a velocity vs. time graph, which is what a person choosing i) Object C was moving in the negative direction at the speed of (-20 m/s). This doesn t make sense since speed is the magnitude of velocity and is always positive. This person much be mixing up velocity and speed. j) Object D traveled 40 m in 2 seconds in the negative direction. This would be true if D was a velocity vs. time graph, which is what a person choosing 7.7 Pose your own problem Answers will vary. In general the main difficulty in this lesson is in reading and labeling their graphs properly. Not realizing the difference between a velocity vs time graph and a position vs time graph will be the cause of most problems. So hint: Stop and say hi to your graph!

7 PUM Physics II - Kinematics Lesson 7 Solutions Page 7 of 7

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