x 2 = (60 m) 2 + (60 m) 2 x 2 = 3600 m m 2 x = m

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3.1 Track Question a) Distance Traveled is 1600 m. This is length of the path that the person took. The displacement is 0 m. The person begins and ends their journey at the same position. They did not change their position after the time interval and so their displacement is 0 m. b) Distance Traveled is 200 m. The distance the person moved is 84.85 m. x 2 = (60 m) 2 + (60 m) 2 x 2 = 3600 m 2 + 3600 m 2 x = 84.85 m 3.2 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. b) Velocity vs. Clock Reading

3.3 Textbook 1. Displacement can be zero even though distance traveled is non zero. A runner that begins a race at the start line, does four labs around the lab, and finishes at the same place they started has traveled a distance of 1 mile but their final position and the initial position are the same and so they have not experienced a displacement. Distance traveled cannot be zero when displacement is non-zero because in order to change your position you must travel some distance. The only way this would be possible is if we could teleport. 2. You can t say anything about their displacement unless you knew the path this person took to get there. This person is referring to the distance they traveled. They could have been walking in a circular path, traveled the distance they said they traveled but ended at the same place they started making their displacement 0 m. 3. You can say that at minimum the object traveled a distance of 300 m. The most direct route to the final position from the initial position would be a straight line between the two points. If the person traveled along this path then the path would be 300 m long, for the displacement to also be 300 m. The person could have taken a longer, less-direct route to the final position and, if they had, the distance traveled would be greater than 300 m. 5. Yes, the average velocity only considers displacement after an amount of time has passed. It does not consider what occures between. The jogger could have jogged at a steady pace, turned around a jogged in the negative direction for a brief moment in time, and then turned again and sprinted towards the positive direction. 3.4 Represent and Reason

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: 3.5 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. a) Bike A started at a positive velocity. Graph A is a velocity vs time graph. At time reading zero Bike A has a velocity of 0 m/s. b) Bike B climbed over a flat hill. Graph B is a position vs time graph, the vertical axis is labeled as position of the object along the coordinate axis. The change in position only occurs in 1-Dimension. c) Bike A stopped twice during the trip. If you consider the bike to be stopped at the zero clock reading because the bike begins with a velocity of 0 m/s then Bike A did stop twice. If you do not, the bike only stops once.

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. e) The last part of the trip bike A was not moving. The line of the velocity vs. time graph is negative. The Bike must be moving in the negative direction. f) The last part of the trip bike B was moving at constant speed in the negative direction. The line of the position vs. time graph is horizontal. The object does not change their position at the end of the observed time interval and so Bike B cannot be moving. 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. 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. 3.6 Represent and Reason 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 b) On this scale it is difficult to notice the slight change in slope from 2-3.3 hrs c) Time (hr)

3.7 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. 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