Question

Meg rowed her boat upstream a distance of 45 mi and then rowed back to the starting point. The total time of the trip was 14 hours. Of the rate of the current was 2 ​mph, find the average speed of the boat relative to the water.

Answer 1

Recall your d = rt.

so hmm is about the same as the other one... so say the boat has a speed rate of "b", well, going downstream the boat goes say " b + 2 ", because the current is adding to it, and going upstream is " b - 2 " because the current is eroding from it.

So if the whole trip took 14 hours, say it took "t" hours going upstream, then going downstream it took the slack or " 14 - t ".

Also notice, the trip forth and back is the same 45 miles.

$\bf \begin{array}{lccclll} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ &------&------&------\\ Upstream&45&b-2&t\\ Downstream&45&b+2&14-t \end{array} \\\\\\ \begin{cases} 45=t(b-2)\implies \frac{45}{b-2}=\boxed{t}\\\\ 45=(b+2)(14-t)\\ ----------\\ 45=(b+2)\left( 14- \boxed{\frac{45}{b-2}}\right) \end{cases} \\\\\\ \cfrac{45}{b+2}=14- {\cfrac{45}{b-2}}\implies \cfrac{45}{b+2}=\cfrac{14(b-2)~-~45}{b-2}$

$\bf \cfrac{45}{b+2}=\cfrac{14b-28-45}{b-2}\implies \cfrac{45}{b+2}=\cfrac{14b-73}{b-2} \\\\\\ 45(b-2)=(b+2)(14b-73)\implies 45b-90=14b^2-45b-146 \\\\\\ 0=14b^2-90b-56\implies 0=7b^2-45b-28 \\\\\\ 0=(7b+4)(b-7)\implies b= \begin{cases} -\frac{4}{7}\\\\ \boxed{7} \end{cases}$