A man can row a distance of 8 km downstream in a certain time and can row 6 km upstream in the same time. If he rows 24 km upstream and the same distance downstream in \(1\frac{3}{4}\) hours, then the speed (in km/h) of the current is:
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This problem involves understanding the concepts of relative speed in the context of boat and stream motion. When a man rows downstream, the speed of the current adds to his speed in still water. When he rows upstream, the speed of the current opposes his speed in still water.
Note that for upstream motion to be possible, the speed of the man in still water must be greater than the speed of the current (\(v_m > v_c\)).
We are given two pieces of information that allow us to set up equations:
Time = Distance / Speed.
Time taken to row 8 km downstream is \(T_1 = \frac{8}{v_m + v_c}\).
Time taken to row 6 km upstream is \(T_2 = \frac{6}{v_m - v_c}\).
Since \(T_1 = T_2\), we have:
\(\frac{8}{v_m + v_c} = \frac{6}{v_m - v_c}\)
Cross-multiplying gives:
\(8(v_m - v_c) = 6(v_m + v_c)\)
\(8v_m - 8v_c = 6v_m + 6v_c\)
Rearranging the terms to one side:
\(8v_m - 6v_m = 6v_c + 8v_c\)
\(2v_m = 14v_c\)
Dividing by 2, we get a relationship between \(v_m\) and \(v_c\):
\(v_m = 7v_c\) --- (Equation 1)
This means the man's speed in still water is 7 times the speed of the current.
The total time taken to row 24 km upstream and 24 km downstream is \(1\frac{3}{4}\) hours, which is \(\frac{7}{4}\) hours.
Time taken to row 24 km upstream = \(\frac{24}{v_m - v_c}\).
Time taken to row 24 km downstream = \(\frac{24}{v_m + v_c}\).
The sum of these times is the total time:
\(\frac{24}{v_m - v_c} + \frac{24}{v_m + v_c} = \frac{7}{4}\) --- (Equation 2)
Now we substitute Equation 1 (\(v_m = 7v_c\)) into Equation 2.
First, find the upstream and downstream speeds in terms of \(v_c\):
Substitute these into Equation 2:
\(\frac{24}{6v_c} + \frac{24}{8v_c} = \frac{7}{4}\)
Simplify the fractions on the left side:
\(\frac{4}{v_c} + \frac{3}{v_c} = \frac{7}{4}\)
Combine the terms on the left side:
\(\frac{4 + 3}{v_c} = \frac{7}{4}\)
\(\frac{7}{v_c} = \frac{7}{4}\)
To find \(v_c\), we can cross-multiply or notice that the numerators are equal, so the denominators must also be equal (assuming \(v_c \neq 0\), which it must be for a current to exist):
\(7 \times 4 = 7 \times v_c\)
\(28 = 7v_c\)
Divide by 7:
\(v_c = \frac{28}{7}\)
\(v_c = 4\)
The speed of the current is 4 km/h.
If \(v_c = 4\) km/h, then \(v_m = 7 \times 4 = 28\) km/h.
Check the first condition: Time for 8 km downstream = \(8/32 = 1/4\) hour. Time for 6 km upstream = \(6/24 = 1/4\) hour. This matches.
Check the second condition: Time for 24 km upstream = \(24/24 = 1\) hour. Time for 24 km downstream = \(24/32 = 3/4\) hour. Total time = \(1 + 3/4 = 7/4 = 1\frac{3}{4}\) hours. This also matches.
The speed of the current is 4 km/h.
| Concept | Formula |
|---|---|
| Downstream Speed | \(v_{down} = v_m + v_c\) |
| Upstream Speed | \(v_{up} = v_m - v_c\) |
| Time | Time = Distance / Speed |
| Term | Description |
|---|---|
| Speed in Still Water (\(v_m\)) | The speed of the boat or man in water that is not flowing. |
| Speed of Current (\(v_c\)) | The speed at which the water is flowing. |
| Downstream | Moving in the same direction as the current. |
| Upstream | Moving against the direction of the current. |
| Relative Speed | The speed of an object relative to another object or a medium. In this case, the boat's speed relative to the ground is affected by the water's speed. |
The problems involving boats and streams are practical applications of the concept of relative speed. When an object moves in a fluid (like water or air) that is also moving, its speed relative to a stationary point on the ground (or outside the fluid) is the vector sum of its speed relative to the fluid and the speed of the fluid itself.
This principle is used not only in boat and stream problems but also in aircraft navigation where wind speed affects the plane's ground speed.
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