Students solve word problems involving rate and distance using a problem solving plan.
After completing this tutorial, you will be able to complete the following:
The distance traveled by an object traveling at a constant speed is given by the formula d = rt, where r is the constant speed, t is the travel time, and d is the distance traveled. This formula, known as the distance formula, can be derived by straightforwardly interpreting speed as a rate of change. Speed measures the rate by which distance changes with respect to time. For example, consider a car traveling at 55 miles per hour. The car moves 55 miles after one hour and an additional 55 miles for every mile thereafter.
The above table calculates distance after one, two, and three hours by adding 55 miles to the total travel distance for every additional hour. This method of determining distance is flawed for three reasons. First, it would be grossly inefficient to use this method to calculate distance for lengthy travel times; a 100-term sum is required to calculate the distance after 100 hours of travel. Second, this method provides no means for determining distance for fractional travel times, such as 1.5 hours. Third, this method provides no insight or predictive ability. These flaws compel us to seek a different method. Consider the table below:
The third column of this table uses multiplication to neatly organize the repeated sum of the second column. The pattern should be evident: the car moves where n is the number of hours of travel time. What then, if a car that travels for hours? We could answer this question by converting the time unit to minutes, but will instead follow a different route by interpreting the speed as a constant rate of change.
The car travels at a constant rate, meaning that its speed does not change from hour to hour, and more importantly, the speed does not change between any two fixed time intervals. On this account, the car travels the same distance in its first 30 minutes of travel as it does during the second, third, or fourth 30 minutes of travel. Therefore, since 30 minutes is one-half hour, the car travels miles for every 30 minutes of travel. Similarly, it travels miles after a quarter hour, or miles after 1/3 of an hour. These calculations resemble formulas derived for whole numbers of hours, given that the car travels hours. With these observations in place, the general formula follows intuitively: the car travels r x t miles after time t (measured here in hours) and speed (measured here in miles per hour). Therefore, the distance formula is d = rt. The distance formula can be rigorously proved by following the framework of this informal argument.
The distance formula can be used to measure the distance between two moving objects. For those comfortable with both absolute values and negative rates, the distance between two moving objects after time t is given by where and are the respective rates of the objects. This formula is independent of the direction of travel and has a strong intuitive basis, but it can obscure an important observation. Namely, if two objects are moving at a constant rate, then the distance between them changes according to the relative speed of one object to the other.
The speed of some object relative to a given reference point is the rate by which the object's distance from the reference point changes over time. For example, the speed of a baseball thrown from a moving vehicle in the direction of travel can be measured from different reference points. If measured from the vehicle, the ball's speed is the rate at which the car moves away from the vehicle. On this account, a pitcher capable of 100 miles per hour pitches would observe that the ball moves at 100 miles per hour. Compare this viewpoint to the stationary observer who, upon measuring the speed of the ball, finds that it is traveling at 100 miles per hour plus the speed of the vehicle. So for example, if the vehicle speeds along at 100 miles per hour, the stationary observer measures the speed as 200 miles per hour.
These sorts of relative speed calculations are accounted for by Galilean relativity, the basic ideas of which should be clear to any airplane passenger who, while impatiently waiting for the food cart, somehow ignores the fact that the food cart actually barrels forward by several hundred miles per hour relative to the Earth.
The distance between two objects moving in opposite directions at constant rates, and units per time interval, increases by units per time interval. For example, if two cars travel in opposite directions, each at 50 miles per hour, then the distance between them increases by 50+50 = 100 miles per hour. This is because each car contributes to the change in distance. Therefore, if the cars start at the same time and place, the distance is given by the formula
If the cars move in the same direction at unequal rates and with the greater, then the distance still increases, albeit at a slower rate. In this situation, the faster car moves away from the slower car, but the rate of increase is lowered by the slower car's speed. Given that the cars start at the same place and time, then the distance is given by Recall that is the faster ratethe slower rate, and t is the travel time. This formula results from observing that the speed of the faster car, relative to the slower, is equal to units per time interval.
These formulas are derivable from the same observations offered as evidence for the usual distance formula,d = rt. This derivation is the main task of the Activity Object.
Please note that to make the problems used in the Activity Object simpler to understand, it is assumed that as soon as the cars start to move, they can immediately achieve their constant rate of travel.
|Approximate Time||40 Minutes|
|Pre-requisite Concepts||Students should know how to solve one-step and two-step linear equations.|
|Type of Tutorial||Problem Solving & Reasoning|
|Key Vocabulary||constant speed, distance, distance formula|