Solving Problems Involving Rate, Time, and DistanceDate: 09/12/2004 at 22:19:09 From: Calvin Subject: How do you calculate time and distance George lives approximately 15 miles from work. He is currently driving 35 mph. Assuming that his speed will remain constant how long will it take George to reach work? I have trouble solving for time with distance and mph. I need to know the formula to make sure I am figuring it right. I know the answer is 25.7 minutes but I do not know how I got it. Date: 09/13/2004 at 09:42:37 From: Doctor Ian Subject: Re: How do you calculate time and distance Hi Calvin, I'll show you two ways to think about these types of problems. First, the standard way and then a more intuitive way. The standard way is to use the formula rate = distance / time If you forget this, you can just think about how we measure rates: miles per hour, feet per second, furlongs per fortnight, and so on. It's always a distance divided by a time. So this is really more of a definition than a formula. Anyway, this formula is really nice if what you have is a distance and a time, and what you want is a rate. But it's not so useful if you have a rate, and either distance or time, and want to get the other. But let's take a little detour for a moment, back to arithmetic. When you first learned about division, you learned that it was just another way of looking at multiplication, e.g., all of the following facts, 3 * 4 = 12 4 = 12 / 3 3 = 12 / 4 are just different ways of expressing the same information. Does that sound familiar? Well, the same is true for the distance-rate-time formula. If r = d / t where r = rate, d = distance, t = time then all of the following, r * t = d r = d / t t = d / r are just different ways of expressing the same information. So you can pick the one that's most useful to you at the moment, i.e., the one where the things you know are on one side, and the thing you want to know is on the other. Now, in your problem, you have a distance (15 miles) and a rate (35 miles per hour). And you want to find a time. So you'd want to choose the version that looks like t = d / r and fill in the values: time = (15 miles) / (35 miles/hour) It's useful to keep the units in there. First, it tells you what the final units have to be: in this case, the miles will cancel, leaving you with hours. Second, in some problems the quantities will intentionally be specified with incompatible units, and you might end up with something like time = (15 miles) / (88 feet/second) In a case like this, you need to convert one of your units so that the measures of distance are compatible. For what it's worth, I haven't bothered to memorize all three versions of the distance-rate-time equation. Whenever I need one, I start with the one I _know_ is going to be true, rate = distance / time and rearrange it as needed. That way, I don't run the risk of remembering it incorrectly. Now, I also mentioned a second way to think about this problem. If George is driving 35 mph, then in one hour he'll go 35 miles; in two hours, 70 miles; and so on. Right? Now, suppose he drives for 15 hours. How far will he go? 35*15 miles. So what fraction of that is the 15 miles he has to go? It's 1/35 of that. So he will be driving for 1/35 of 15 hours: 15 hours 15 hours * 60 min/hr -------- = -------------------- 35 35 = about 25.7 minutes Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/