Question 180490
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This is a distance, rate, and time problem, so the basic formula to use is *[tex \Large d = rt].  However, we will find it convenient to express this relationship as *[tex \Large t = \frac {d}{r}] for this particular problem.


We are asked to find the rate of the boat in still water, so let's call that *[tex \Large r_s] to distinguish it from the <i>r</i> in the basic formula above.  We are given the speed of the current, 5 mph, so we know that the speed of the boat relative to the shoreline as the boat goes upstream is *[tex \Large r_s - 5] and the speed of the boat relative to the shoreline as the boat goes downstream is *[tex \Large r_s + 5].  We also know that each leg of the trip is 40 miles long.  Lastly, let's say that the time it takes for the boat to make the upstream trip is <i>t</i> hours.  That means that, since the entire trip took 6 hours, the downstream trip must have taken 6 - <i>t</i> hours.


Now let's use the formula, *[tex \Large t = \frac {d}{r}] to describe the upstream trip:


*[tex \Large \text {          }\math t = \frac {40}{r_s - 5} ]


And the downstream trip:


*[tex \Large \text {          }\math 6 - t = \frac {40}{r_s + 5}]


First, solve the downstream equation for <i>t</i>:


*[tex \Large \text {          }\math t = 6 - \left(\frac {40}{r_s + 5}\right)]


And simplify the result:


*[tex \Large \text {          }\math t = \frac {\[6(r_s + 5)\] - 40}{r_s + 5}]


*[tex \Large \text {          }\math t = \frac {6r_s - 10}{r_s + 5}]


Next if you compare the simplified form of the downstream equation with the upstream equation, you will notice that we have two different expressions in *[tex \Large r_s] that are both equal to the same thing, namely <i>t</i>.  We can now set these two expressions equal to each other:


*[tex \Large \text {          }\math \frac {40}{r_s - 5} = \frac {6r_s - 10}{r_s + 5} ]


Put everything on the left and apply the common denominator *[tex \Large {r_s}^2 - 25]


*[tex \Large \text {          }\math \frac {\[40(r_s + 5)\] - \[(6r_s - 10)(r_s - 5)\]}{{r_s}^2 - 25} = 0 ]


Multiply by *[tex \Large {r_s}^2 - 25]:


*[tex \Large \text {          }\math \[40(r_s + 5)\] - \[(6r_s - 10)(r_s - 5)\] = 0 ]


Apply the distributive property and collect terms:


*[tex \Large \text {          }\math 40r_s + 200 - (6{r_s}^2 - 30r_s - 10r_s + 50) = 0 ]


*[tex \Large \text {          }\math -6{r_s}^2 + 80r_s + 150 = 0 ]


Multiply by *[tex \Large -\frac {1}{2}]


*[tex \Large \text {          }\math 3{r_s}^2 - 40r_s - 75 = 0 ]


Factor:


*[tex \Large \text {          }\math (3r_s + 5)(r_s - 15) = 0 ]


Therefore *[tex \Large \text {          }\math r_s = -\frac {5}{3} ] or *[tex \Large \text {          }\math r_s = 15 ]


The negative answer is absurd because we know the boat wasn't going backwards, so exclude that answer as an extraneous root introduced by the action of squaring the variable during the solution process, and that leaves us with  *[tex \Large \text {          }\math r_s = 15 ]


Check the answer.


If the speed in still water is 15 mph, then the upstream speed must have been 15 - 5 = 10 mph.  At 10 mph, it takes *[tex \Large 40 \div 10 = 4] hours to make the trip.  The downstream speed must have been 15 + 5 = 20 mph.  At 20 mph, it takes *[tex \Large 40 \div 20 = 2] hours to make the trip.  4 hours plus 2 hours = 6 hours which was the given elapsed time.  Answer checks.





John
*[tex \LARGE e^{i\pi} + 1 = 0]
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