Several Ways to Solve a Rate-Time-Distance ProblemDate: 02/14/2006 at 01:17:41 From: Al Subject: When and where will the car pass the bus? A bus leaves point A travelling at a speed of 10 mph. 22 minutes later a car leaves point A travelling in the same direction at 40 mph. At what distance will the car and the bus meet, and how much time will have elapsed? Is there a formula for a problem like this? Date: 02/14/2006 at 10:02:53 From: Doctor Greenie Subject: Re: When and where will the car pass the bus? Hi, Al -- The basic formula is the familiar distance = rate * time but few math problems are solved simply by applying formulas. We need to find a way to use the formula to solve our particular problem. I see several ways to approach your problem. Let me show you three of them; you can see which of them makes most sense to you and then you can use that method the next time you run into a problem like this. Or better yet, try to understand all three methods so you can choose to apply any one of them to a particular problem--because any one of the approaches might be obviously easier to apply to any particular problem. Probably the traditional approach to the problem is using algebra and the basic distance/rate/time formula. To begin with, we need to remember to keep our units straight--because the speeds are given in mph and the time difference is given in minutes. 22 minutes is 22/60 of an hour, so we can say let t = time car travels (hours) then t + 22/60 = time bus travels (hours) We have defined expressions for the times the two vehicles travel; and the problem tells us the two speeds. So our basic formula distance = rate * time tells us 40t = distance car travels 10(t + 22/60) = distance bus travels We want to know when the distances traveled by the two vehicles are the same because when the car catches the bus they will each have gone the same distance from point A, so we can write and solve an equation which says just that: 40t = 10(t + 22/60) Solving this equation is then basic algebra: 40t = 10t + 22/6 30t = 22/6 t = 22/180 = (7 1/3)/60 (hours) = 7 1/3 minutes A different use of the basic formula to solve the problem involves finding how far the bus has gone before the car starts and then using the DIFFERENCE of the speeds to determine how long it will take the car to catch up to the bus. The thinking behind this approach is that, after the car starts, the bus is still moving, so the car is CATCHING UP TO the bus at a rate equal to the DIFFERENCE between the two speeds. The distance the bus travels before the car starts is 10(22/60) = 22/6 miles The car is catching up to the bus at the rate of 30mph. To find out how long it takes for the car to catch up, we use the following variation of the basic formula: time = distance/rate t = (22/6)/30 = 22/180 (hours) = ... = 7 1/3 minutes The third method I easily see for solving this problem uses the ratios of the speeds instead of the speeds themselves. The thinking behind this approach is that, since the car travels 4 times as fast as the bus, the bus will take 4 times as long as the car to go a particular distance. Again (as in the first method above) we start with let t = time car travels (hours) then t + 22/60 = time bus travels (hours) Then, since we know the bus takes 4 times as long to go a particular distance (since the car at 40 mph is 4 times as fast as the bus at 10 mph), we have (t + 22/60) 4 ----------- = - t 1 4t = t + 22/60 3t = 22/60 t = 22/180 (hours) = ... = 7 1/3 minutes I hope I haven't confused you by showing several different approaches to your problem. Different readers are going to find different ones of these methods easier to understand and to use; I like to give them that choice. And I personally like to have different methods to choose from. One method or another might be easier to apply to any particular problem. Or I might find for some reason that I can't get a particular method to work out, and I can then use another method to solve the problem. Or I can even get a confidence check in my answer if I can solve a particular problem by more than one method and get the same answer. I hope all of this helps. Please write back if you have any further questions about any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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