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### Solving Problems Involving Rate, Time, and Distance

```Date: 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
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.

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

Does this help?

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```
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