systemt3

Solving a word problem using a system of linear equations: Problem type 3

These problems involve uniform motion, usually in wind or water.

For any uniform motion the equation rate times time equal distance holds.

rate x time = distance where rate is commonly called speed.

When traveling in water, going with the current will add to the rate in still water (no current). Traveling against the current (upstream) will be slower. The rate will be the still water rate minus the rate of the current.

Upstream rate = rate in still water (r) - rate of current (c) = r - c

Downstream rate = rate in still water (r) + rate of current (c) = r + c

Similarly, when one travels in wind the rate is either increased or decreased depending on whether one is traveling with the wind (tailwind) (r + w) or against the wind (headwind)(r - w)

It is often helpful to organize information from these types of problems in a table. Headings on the columns will be the rate * time = distance formula. Once you fill in the given information and use variables for the unknown cells in the table, you can multiply across each row to get two equations.

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Example1:

A motorboat travels 531 km in 9 hours going upstream and 428 km in 4 hours going downstream. What is the speed of the boat in still water, and what is the speed of the current?

Our two equations will come from the two trips: one upstream, the other downstream. Let r be the rate of the boat in still water and c represent the rate of the current.

	Rate*	Time	=Distance
Upstream	r - c	9	531
Downstream	r + c	4	428

Then (rate going up stream) times (time traveled) = distance traveled. The first row gives us the following equation:

(r - c) 9 = 531

Also, (rate going downstream) times (time traveled) = distance traveled. The second row gives us this equation:

(r + c) 4 = 428

Using the distributive property to get rid of the parentheses gives the equations:

9r - 9c = 531

4r + 4c = 428

Solving this system using elimination (addition method) gives r = 83 km per hour and c = 24 km per hour.

Thus, the speed rate of the boat in still water is 83 km per hour,
and the speed rate of the current is 24 km per hour.

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Example 2:

Solving a word problem using a system of linear equations: Problem type 3

A child in an airport is able to cover 348 meters in 3 minutes running at a steady speed down a moving sidewalk in the direction of the sidewalk's motion. Running at the same speed in the direction opposite to the sidewalk's movement, the child is able to cover 312 meters in 4 minutes. What is the child's running speed on a still sidewalk, and what is the speed of the moving sidewalk?

This problem is very much like the airplane flying against/with the wind type problem and the boat rowing with or against the current. In these problems the rate isn’t simply how fast you can walk (or row, or fly). Rate is affected by the speed of the wind, the speed of the current, or in this problem the speed of the moving sidewalk.

Thus, in these types of problems it takes two elements to represent the rate…the speed of the actual object (plane, child, boat, etc.) and the speed of whatever the object is moving in (air, sidewalk, water, etc.). The actual speed of the object when it is moving in the same direction as the wind, water, sidewalk, etc. is the sum of the rate it could move without the medium plus the rate the medium moves.

In the sidewalk problem above, if we let c represent the rate the child walks and s represent the rate the sidewalk moves, then c + s represents how fast the child moves when walking in the same direction as the sidewalk. Filling the information into our rate times time equals distance formula, rt=d) gives:

	Rate*	Time	=Distance
With the sidewalk	r + c	3	348
opposite the sidewalk	r - c	4	312

So our first equation (from row 1 in our table) is

(c + s)*3 = 348. We could distribute the 3, or we could divide both sides by 3 to get the simpler equation

c + s = 116

When the child moves the opposite way as the side walk, the actual rate will be how fast the child can walk minus how fast the sidewalk is moving in the opposite direction. The second row of the table gives the equation:

(c – s)4 = 312m per s. Again, we could distribute the 4 to get rid of the parentheses, or we can divide both sides by 4 to get the simpler equation:

c - s = 78

Once we have the two equations, they are easy to solve by elimination (adding them together to eliminate the s variable)

c + s = 116

c – s = 78

So, 2c = 194.

Thus, c = 97 m/s

The child can walk 97 m/s. Substituting that value back into either equation and solving for s gives us the sidewalk rate of 19 m/s.