Speed and Velocity Speed refers to how far an object travels in a given amount of time, regardless of direction. If a car travels 100 km in 2 hours, it s average speed was 50km/hour. 100km = 50 km/hr 2 hours The average speed of an object is the total distance traveled divided by the total time it takes to travel this distance. If a caterpillar travels 3 inches in 1 hour, then 12 inches in 2 more hours, the average speed was 5 inches/hour. Total Distance = 3 inches + 12 inches = 15 inches Total Time = 1 hour + 2 hours = 3 hours 15 inches 3 hours = 5 inches/hr Speed is always a positive number. Velocity signifies the speed of an object AND the direction it is moving. The average velocity is the change in position (called displacement) divided by the total time. If an object moves 20 meters east, then 5 meters west, the change in position is 15 meters east. 20 meters east + 5 meters west = change in position of 15 meters east If the total travel time was 5 minutes, the average velocity is 15m /5 minutes = 3 meters/minute, east The speed of this example is 25m/5 minutes = 5 meters/minute Another way to describe the above example is like this: +20 meters -5 meters = +15 meters TO SUMMARIZE: Speed is distance over time. Speed does not depend on direction. Velocity is the change in position over time. Velocity depends on direction.
Velocities, using vectors (arrows that show the amount and direction) Canoe moves: River current: Net velocity: 10 m/sec + 10 m/sec = 20 m/sec -10 m/sec + -10 m/sec = -20 m/sec 10 m/sec + -10 m/sec = 0 m/sec OR 10 m/sec east + 10 m/sec east = 20 m/sec east 10 m/sec south + 10m/sec north = 0 m/sec Canoe moves: River current: Net velocity: 10 m/sec east + 10m/sec north = 14.1 m/sec northeast 10m/sec NW + 10m/sec NE = 10 m/sec North! 30 30
Speed and Velocity Examples: 1. A person walks 4 meters East, 2 meters South, 4 meters West, and finally 2 meters North. The entire motion lasted for 24 seconds. Determine the average speed and the average velocity. Avg. Speed: 12 meters/24 seconds = 0.5 m/s. Avg. Velocity: change in position = zero so V= 0m/24seconds = 0m/s 2. A ball is tossed into the air with the following data. The ball was 6 feet above the ground when released, 90 feet after 1 second, etc. Adapted from http://www.calculusapplets.com/avevel.html Time (sec) 0 1 2 3 4 5 6 height (feet) 6 90 142 162 150 106 30 What was the velocity? Let s say up = positive velocity and down = negative velocity. It depends on the time interval Over the first second: D = 90-6 =84 ft. T = 1 second V = 84 ft/sec Over the 6th second: D = 30-106 = -76 ft. T = 1 second V = -76ft/sec The average velocity for the entire interval: D = 30-6 ft = 24 ft. T = 6 seconds V = 4 ft/sec The average speed for the entire interval: D = lowest height (6ft) to heighest height (162ft) = 156 feet PLUS heighest height (162ft) to final height (30ft) = 132ft. 156 + 132 = 288 ft T = 6 seconds S = 288 ft / 6 seconds = 48 ft/sec Written as an equation: (162ft 6ft) + (162ft 30ft) = 6 seconds
Speed and Velocity Problems: 1. A skateboarder zig zags back and forth, going 3 meters north, then 4 meters south, then 3 meters north, then 5 meters south. This takes 1 minute. What is the skateboarder s average speed and average velocity? (north = + south = ) 2. Use the diagram to determine the average speed and the average velocity of the skier during these three minutes. 50 3. A football coach paces back and forth along the sidelines. The diagram below shows several of coach's positions at various times. At each marked position, the coach makes a "U-turn" and moves in the opposite direction. In other words, the coach moves from position A to B to C to D. What is the coach's average speed and average velocity? 4. A car travels 25 km/hr North for 4 hours, then at 50 km/hr North for 8 hours, and finally 20 km/hr North for 2 hours. Find (a) the total distance covered (b) the average speed, and (c) the average velocity for the complete trip. 5. A train travels at an average velocity of 120 km/h for 3 hours, then at 80 km/h for 210 km. The train stops for 1 hour. Finally, the train covers 300 km in 2.9 hours. All motion is to the NORTH. (a) What is the train's displacement at the end of the first 3 hours? (b) How much time was spent going 80 km/h in the second leg of the trip? (c) What was the average velocity for the final 3.9 hours of the trip? (d) Calculate the average velocity for the whole trip. (e) Calculate the average velocity if the second part of the trip (80 km/h for 210 km) were to the SOUTH.
1. Answer avg. speed = 15 meters/1 minute = 15 meters/minute avg. velocity = (+3 + -4 + +3 + -5) = -3m/1 minute = -3 m/min 2. Answer minute 1 = +190 meters minute 2 = - 140 meters minute 3 = + 100 meters avg. speed = 430 m / 3 minutes = 143 m/min avg. velocity = +150 m / 3 minutes = + 46.7 m/min 3. Answer 0-3 minutes (3 min) = -35 yards 3-6 minutes (3min) = + 20 yards 6-10 minutes (4 min) = - 40 yards avg. speed = 95 yards / 10 minutes = 9.5 yds/min avg. velocity = -55 yards / 10 minutes = -5.5 yds/min 4. Answer: 25 km/hr North for 4 hours = 100km for 4 hours 50 km/hr North for 8 hours = 400km for 8 hours 20 km/hr North for 2 hours = 40km for 2 hours Total distance = 540km Total time = 14 hours Average velocity = 540/14 = 38.6 km/hour north 5. Answers: a) 120 x 3 = 360 km N b) 210/80 = 2.6 hr c) 300/3.9 = 76.9 km/hr N d) D = 360 + 210 + 0 + 300 =870 km north T = 3 + 2.6 + 1 + 2.9 = 9.5 hours V = 870/9.5 = 91.6 km/hr north e) D = 360 210 + 0 + 300 = 450 km north V = 450/9.5 = 47.4 km/hr N
Resources for this document: Giancoli.Physics. Pearson/Prentice Hall. 2005 http://www.glenbrook.k12.il.us/gbssci/phys/class/1dkin/u1l1d.html http://www.glenbrook.k12.il.us/gbssci/phys/class/vectors/u3l3a.html Cherniak doc s