Question

A motorboat is capable of traveling at a speed of 14 miles per hour in still water. On a particular day, it took 15 minutes longer to travel a distance of 12 miles
upstream than it took to travel the same distance downstream. What was the rate of current in the stream on that day?

Answer 1

By solving a system of equations we will find that the rate of current in the stream is S = 2 mi/h.

When the motorboat travels downstream, the total velocity will be the velocity of the motorboat in still water plus the velocity of the stream, while if the motorboat travels upstream, we have the velocity of the stream subtracted.

So upstream the speed is:

(14 mi/h - S)

Downstream the speed is:

(14 mi/h + S)

Where S is the rate of current in the stream.

We know that downstream it takes 15 minutes more to travel 12 miles, then we can write the system of equations:

(14 mi/h + S)*T = 12 mi

(14 mi/h - S)*(T - 15 min) = 12 mi

To solve this, we need to isolate one of the variables in one of the equations, I will isolate T in the first one:

T = (12 mi)/(14 mi/h + S)

Replacing that in the other equation we will get:

(14 mi/h - S)*((12 mi)/(14 mi/h + S) - 15 min) = 12 mi

Now we can solve this for S. Now we can multiply both sides by (14 mi/h + S).

(14 mi/h - S)*12 mi  - (14 mi/h + S)*(14 mi/h - S)*(- 15 min) = 12 mi*(14 mi/h + S)

Also notice that the speeds are in hours, so we can rewrite:

- 15 min = -0.25 h

(14 mi/h - S)*12 mi  - (14 mi/h + S)*(14 mi/h - S)*(- 0.25 h) = 12 mi*(14 mi/h + S)

168 mi^2/h - 12mi*S  + 49mi^2/h + 0.25h*S^2 = 168mi^2/h + 12mi*S

- 12mi*S  + 49mi^2/h - 0.25h*S^2 = 12mi*S

-24mi*S -  0.25h*S^2  + 49mi^2/h = 0

This is a quadratic equation, the solutions are:

$S = \frac{24mi \pm \sqrt{(-24mi)^2 - 4*(49mi^2/h)*(-0.25h)} }{2*-0.25h} \\\\S = \frac{24mi \pm 25 mi }{-0.5h}$

We only take the positive solution, so we get:

S = (24 mi - 25 mi)/(-0.5 mi) = 2 mi/h

The rate of current in the stream is 2 mi/h.

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