In the second team, two identical twins always run at the same speed. In one of the teams, the speed of the first athlete is 20 mph, that of the second is 30 mph. Let there be a relay race between two teams each consisting of two athletes. There is another way to look at the situation. What we found is that, depending on the circumstances, to determine the average speed of motion, computations based on the same basic formula (S = D/T) may have to follow different routes. If the distance between the cities were, as in the first problem, 50 miles, then the journey would take T = 50/24 hours or 2 hours and 5 minutes, from which 1 hour and 15 minutes (= 25/20 hours) were spent on the first 25 miles and 50 minutes (= 25/30 hours) on the second 25 mile stretch. It follows that his average speed is given by S = 2d/(d/20 + d/30) or, after cancelling out the common factor d, S = 2/(1/20 + 1/30) = 24 mph. During that time he covered D = 2d miles. Therefore, on the whole, the fellow was on the road T = d/20 + d/30 hours. The first leg of the journey took d/20 hours, the second d/30. Let d be half the distance between the two cities. There are three ways to write the formula above: S = D/T, D = ST, T = D/S. The second problem is only a little more complex. The average speed then is found to be S = 50/2 = 25 mph. Therefore, in 2 hours he traveled the total of D = 20 + 30 = 50 miles. Similarly, in the second hour he covered the distance of 30 miles. Going at 20 mi/h for the first hour the fellow covered the distance of 20 miles. The fellow was on the road for the total of T = 2 hours. The definition applies directly to the first problem. The second one, as a mathematical conundrum, has been included in many math puzzles books.īy definition, the average speed S of the motion that lasted time T over the distance D isĪverage speed = (Total distance)/(Total time) or S = D/T The two problems are often confused and the difference between them may not be immediately obvious. Then he (instantaneously) increased his speed and traveled the remaining distance at 30 miles per hour. The first half of the way, he drove at the constant speed of 20 miles per hour. Find the average speed of the motion.Ī fellow travels from city A to city B. Then he (instantaneously) increased his speed and, for the next hour, kept it at 30 miles per hour. For the first hour, he drove at the constant speed of 20 miles per hour. Feels great.Consider two similar problems ( there is a more formal discussion): Problem 1Ī fellow travels from city A to city B. Next time you are out walking, imagine you are still and it is the world that moves under your feet. Think about this: are you really standing still? You are on planet Earth which is spinning at 40,075 km per day (about 1675 km/h or 465 m/s), and moving around the Sun at about 100,000 km/h, which is itself moving through the Galaxy. When we say something is "at rest" or "moving at 4 m/s" we forget to say "in relation to me" or "in relation to the ground", etc. Yes, the velocity is zero as you ended up where you started. Velocity = 130 m 100 s East = 1.3 m/s East You forgot your money so you turn around and go back home in 120 more seconds: what is your round-trip speed and velocity? Example: You walk from home to the shop in 100 seconds, what is your speed and what is your velocity?
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