Question

A motorboat can maintain a constant speed of 37 miles per hour relative to the water. The boat makes a trip upstream to a certain point in 47 minutes; the return trip takes 27 minutes. What is the speed of the current?

Answer 1

We have to find the speed of the current (Vc).

We now that a trip upstream took 47 minutes (47/60 hours, with the current against the movement of the boat) and a trip downstream took 27 minutes (27/60 hours).

The speed of the boat is constant with a value of Vb = 37 miles per hour.

As the average speed is equal to the distance divided by the time, we can write for the trip upstream:

$V_b-V_c=\frac{d}{t}=\frac{d}{47\/60}=\frac{60}{47}d$

For the trip downstram the current is in favor of the boat so we can write:

$V_b+V_c=\frac{d}{t}=\frac{d}{27\/60}=\frac{60}{27}d$

We can then add the two equations as:

$\begin{gathered} V_b-V_c+(V_b+V_c)=\frac{60}{47}d+\frac{60}{27}d \\ 2V_b=60d(\frac{1}{47}+\frac{1}{27}) \\ 2\cdot37=60d\cdot(\frac{27+47}{47\cdot27}) \\ 74=60d(\frac{74}{1269}) \\ d=\frac{74\cdot1269}{74\cdot60} \\ d=21.15\text{ }miles \end{gathered}$

Knowing the distance d = 21.15 miles, we can find the speed of the current as:

$\begin{gathered} V_b+V_c=\frac{60d}{27} \\ V_c=\frac{60d}{27}-V_b \\ V_c=\frac{60\cdot21.15}{27}-37 \\ V_c=47-37 \\ V_c=10\text{ }mph \end{gathered}$

Answer: the speed of the current is 10 miles per hour.