Intermediate algebra

[Matthew1]

Problem
boat speed in still water is 15mph
If boat runs against the current it goes 9 miles in the same time period as below
If boat runs with the current it goes 12 miles in same time period
What is the speed of the current?
(My grandson's problem which I cannot solve - - - embarassed grandpa)

 

[Loren]

Problem
boat speed in still water is 15mph
If boat runs against the current it goes 9 miles in the same time period as below
If boat runs with the current it goes 12 miles in same time period
What is the speed of the current?
(My grandson's problem which I cannot solve - - - embarassed grandpa)

Name stuff.
Rate of boat in still water = 15 mph.
Rate of water = x
Rate of boat going downstream = 15+x
Rate of boat going upstream = 15-x

Rate X Time = Dist
15+x....t.......12 <<<Downstream
15-x....t.......9 <<<Upstream

Can you take it from here?

 

[mmm4444bot]

… embarassed grandpa …

Probably not your fault (or your grandson's, either, if he's embarrased, too).

Time and time again, I've seen instructors assign this exercise without saying a single word about physics.

We can think of the current as a force that speeds up objects going with the flow by pushing them along.

We can think of the current as a force that hinders object going against the flow by pushing against them.

That's why the speed of the boat is increased by the speed of the current x, and reduced by the speed of the current x, accordingly.

If you actually had this physics concept in hand, already, then perhaps your grandson's issue is writing a final equation to solve for x. Show him what Loren posted, and see if he gets it.

If not, then show him the following similar example.

If he still needs more help, then please show us what he is able to accomplish or tell us why he's stuck.



We always begin with the relationship between how fast something moves, the speed at which it moves, and the amount of elapsed time during which it moves as follows.

Distance = Rate * Time

If the rate of some object is expressed as X + 50, and the object moves at that constant rate for some unknown period of elapsed time (expressed as T), then the distance covered during that time period is:

Distance = (X + 50) * T

Likewise, if the rate going in the opposite direction is expressed as X - 50, and the object moves at this reduced rate for the same amount of elapsed time, then the distance covered in the opposite direction is:

Distance = (X - 50) * T

Let's say that these two distances are 100 miles and 75 miles, respectively.

So, in this senario, the object moves 100 miles at the faster rate during the same amount of time that it takes to go 75 miles at the slower rate. I was going to keep it generalized, but, heck, let's say that the object is a spy drone that flies steadily at 50 miles per hour in still air. The symbol X represents wind speed (a force that speeds up the drone when flying with the wind, and impedes it when flying against the wind). So, we're trying to find out how windy it is.

So, our two equations for Distance = Rate * Time become:

(X + 50) * T = 100

(X - 50) * T = 75

Now, it's the fact that the elapsed time is the same in both directions that allows us to get a single equation to solve for x.

Going back to the basic relationship, we can get an expression for Time by dividing both sides by Rate:

Distance = Rate * Time

Distance/Rate = (Rate * Time)/Rate

Distance/Rate = Time

In other words, just as we can express some distance as the product of a rate and time, we can express elapsed time as a ratio of distance over rate.

I'm now going to do this with the two equations above.

T = 100/(X + 50)

T = 75/(X - 50)

Because the elapsed time is the same in both directions, both of these expressions for T are equal.

100/(X + 50) = 75/(X - 50)

Now we have a single equation containing the symbol X, so we can solve it. Cross-multiplying clears the fractions.

100(X - 50) = 75(X + 50)

By multiplying out both sides using the Distributive Property, then combining like terms, and finally isolating X on one side of the equals sign, we get the windspeed X.

I'm not typing out these last few steps because your grandson likely knows how to do them.

In this example, X turns out to be 350. In other words, the wind is blowing at 350 miles per hour. (Yikes. I pulled the numbers out of thin air, for this example, so that's why the result is unrealistic. Plus, I'm enjoying my wine, right now. Anyhoo, it's the strategy that's important; not the value 350.)

(If your grandson cannot solve this equation for X, then he might not be ready yet to work on these types of word problems. Let us know, and we can provide links to lessons on solving basic equations containing a single symbol.)