Word Problem

(Distances)

Instructor: Dr. Jo Steig

This first example is fairly typical of the steps we will use to solve a distance problem and it will serve as a template for subsequent problems of this type.

Example 1: Jan and John are going to drive their motorcycles from San Francisco to San Diego. Jan leaves at 8:00 am but John is unable to leave until 8:30 am. If Jan drives at 50 mph and John drives at 55mph, how long before John will catch up with Jan?

Step 1: Identify which column you can complete from the information given in the problem without having to use a variable and fill it out. In this case, we know the rate that both Jan & John are going so we will fill that part in.

rate X (time) = distance

Jan 50 mph

John 55 mph

Step 2: Identify what you are looking for and define your variable. Here we are looking for the number of hours before John catches up with Jan
Let t represent the number of hours John drives to catch up with Jan.

Place t in the appropriate cell (John's drive time)

Fill out the rest of the column using the relationship between John's drive time and Jan's drive time. That is, Jan started .5 hour before John so her time will be (t + .5)

rate X (time) = distance

Jan 50 mph t + .5

John 55 mph t

Step 3: Fill out the last column by using the first two columns and following the equation at the top of the chart (rate X time = distance).

rate X (time) = distance

Jan 50 mph t + .5 50 (t + .5)

John 55 mph t 55 (t )

Step 4: The equation that will model this problem will now be drawn from the last column (distance). Go back to the original problem and determine how the distance Jan travels and the distance John travels are related. In this case, since they both begin in San Francisco, take the same route, and John finally overtakes Jan, they travel the same distance. So, Jan's distance = John's distance....or
50 (t + .5) = 55 (t )

Step 5: Now solve the equation and answer the original question.

Solving, we get: t = 5

Answer to the original question: John will take 5 hours to overtake Jan.

The next example is an illustration of how different situations (resulting in different equations) can be constructed from this basic scenerio.

Example 2: Jan and John are going to drive their motorcycles from San Francisco to San Diego. Jan leaves at 8:00 am but John is unable to leave until 8:30 am. If Jan drives at 50 mph and John drives at 55mph, how long before John will be 25 miles ahead of Jan?

Notice how similar this problem is to the first example. In fact, the only way this problem differs from the original one is in the relationship between the two distances traveled by John and Jan.
In this problem, John ultimately travels 25 miles more than Jan travels. The modeling equation will again be drawn from the distance column of the table, but will reflect this new relationship. For convenience, I'll repeat the table from the last problem.

Once again, t represents John's rate.

rate X (time) = distance

Jan 50 mph t + .5 50 (t + .5)

John 55 mph t 55 (t )

Remember, John's distance is 25 more than Jan's distance.

Translating this into an algebra statement:

55 (t ) = 25 + 50 (t + .5)
Note: We add 25 to Jan's distance because her distance is smaller than John's. We add 25 to the smaller distance to make the two sides of the equation equal.

Now solve the equation and answer the original question.

Solving we get: t = 10

Answer to the original question: It will take John 10 hours before he is 25 miles ahead of Jan.

Before we leave John and Jan, let's try one more scenerio.

Example 3: Jan and John are 500 miles apart and are driving their motorcycles toward on another. Jan leaves at 8:00 am but John is unable to leave until 8:30 am. If Jan drives at 50 mph and John drives at 55mph, how long before John and Jan meet?

Once again, the only way this problem differs from the original one is in the relationship between the two distances traveled by John and Jan.
In this case, it is the SUM of their distances that equals 500 (since they are 500 miles apart and eventually meet. The table stays the same and we draw our modeling equation from the distance column.

Once again, t represents John's rate.

rate X (time) = distance

Jan 50 mph t + .5 50 (t + .5)

John 55 mph t 55 (t )

John's distance added to Jan's distance is equal to 500 miles.
Translating this into an algebra statement:

55 (t ) + 50 (t + .5) = 500

Solving we get: t = 472/105 or approximately 4.52

Answer to the original question: John and Jan will meet approximately 4 hours and 31minutes after John begins to drive.

If you have other types of distance problems that you would like to see here, please email me and I'll see what I can do to accommodate your request.

© 1999 Jo Steig

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