Chapter 6.4, Problem 18PPE

To determine

To find: the tugboat’s speed in still water and the speed of water current.

Answer to Problem 18PPE

Tugboat’s speed in still water = 10 mi/h

Speed of the water current = 2 mi/h

Explanation of Solution

Given information:

A tugboat can pull a boat 24 mi downstream in 2 h. Going upstream, the tugboat can pull the same boat 16 mi in 2 h.

The tugboat’s speed going downstream is $=\frac{24}{2}=12\text{ mi/h}$

The tugboat’s speed going upstream is $=\frac{16}{2}=8\text{ mi/h}$

Now, let the tugboat’s speed in still water be $x$

And let the speed of water current be $y$ .

Then

  $\begin{array}{l}x+y=12\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{.(1)}\\ x-y=8\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (2)}\end{array}$

Adding equation (1) and (2) we get,

  $\begin{array}{l}x+y+x-y=8+12\\ \Rightarrow 2x=20\\ \Rightarrow x=\frac{20}{2}=10\end{array}$

Putting $x=10$ in equation (1) we get,

  $\begin{array}{l}\text{10}+y=12\\ \Rightarrow y=12-10=2\end{array}$

Thus, tugboat’s speed in still water = 10 mi/h

Speed of the water current = 2 mi/h

The tugboat’s speed going downstream is 12 mi/h which is 2mi/h greater than the tugboat’s speed in still water and it is 6 times the speed of water current.

The tugboat’s speed going upstream is 8 mi/h which is 2mi/h less than the tugboat’s speed in still water and it is 4 times the speed of water current.