Question

A tugboat goes 180 mils upstream in 10 hours. The return trip downstream takes 5 hours. Find the speed of the tugboat without a current and the speed of the current.

The speed of the tugboat is ______ mph, and the speed of the current is _______mph. (Simplify your answers).

Answer 1

We know:
\[
\begin{align}
\frac{180}{10}&=18\text{ mph}\\
\frac{180}{5}&=36\text{ mph}
\end{align}
\]Since the current is constant, then, we know it goes:
\[
\frac{36-18}{2}=9\text{ mph}
\]Therefore the speed of the tugboat is:
\[
\frac{18+36}{2}=18+9=36-9=27\text{ mph}
\]Without a current. And, by this, we are done.

Answer 2

@LolWolf Thank you for not just giving me the answer but showing the work. It helps me so that I am able to solve the problem myself next time. Thank you, thank you thank you. One more question. How do I know when to divide?

Answer 3

Sure thing. And, we know that the speed of the boat is the average of the speeds going upstream and downstream. (i.e. if the boat goes at a speed \(v\) and the river goes at a speed \(r\), then, it goes up-stream at a speed \(u=v-r\) and downstream at a speed \(d=v+r\), which means, that, solving for \(v\):
\[
\frac{d+u}{2}=\frac{v+r+v-r}{2}=\frac{v+v+r-r}{2}=\frac{2v}{2}=v
\]And the river's speed is:
\[
\frac{d-u}{2}=\frac{v+r-(v-r)}{2}=\frac{v+r-v+r}{2}=\frac{v-v+r+r}{2}=\frac{2r}{2}=r
\]Since we are given \(d\) and \(u\) we use the formulas to show this.

Answer 4

Thank you!!!