Problem Analysis:
- Boat speed in still water (\(v_b\)) = 8 km/hr.
- Total time for the journey = 4 hours.
- Distance upstream = 12 km.
- Distance downstream = 18 km.
- Goal: Find the speed of the stream (\(v_s\)).
Calculations for Stream Speed
Let the speed of the stream be \(v_s\) km/hr.
- Upstream speed (\(v_u\)) = Speed of boat - Speed of stream = \((8 - v_s)\) km/hr.
- Downstream speed (\(v_d\)) = Speed of boat + Speed of stream = \((8 + v_s)\) km/hr.
The time taken is calculated using the formula: Time = Distance / Speed.
- Time upstream = \(\frac{12}{8 - v_s}\) hours.
- Time downstream = \(\frac{18}{8 + v_s}\) hours.
The total time is the sum of upstream and downstream times:
\( \frac{12}{8 - v_s} + \frac{18}{8 + v_s} = 4 \)Simplify the equation by dividing by 2:
\( \frac{6}{8 - v_s} + \frac{9}{8 + v_s} = 2 \)Combine the fractions:
\( \frac{6(8 + v_s) + 9(8 - v_s)}{(8 - v_s)(8 + v_s)} = 2 \) \( \frac{48 + 6v_s + 72 - 9v_s}{64 - v_s^2} = 2 \) \( \frac{120 - 3v_s}{64 - v_s^2} = 2 \)Cross-multiply:
\( 120 - 3v_s = 2(64 - v_s^2) \) \( 120 - 3v_s = 128 - 2v_s^2 \)Rearrange into a quadratic equation (\(ax^2 + bx + c = 0\)):
\( 2v_s^2 - 3v_s + 120 - 128 = 0 \) \( 2v_s^2 - 3v_s - 8 = 0 \)Solve using the quadratic formula \(v_s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=2\), \(b=-3\), \(c=-8\):
\( v_s = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-8)}}{2(2)} \) \( v_s = \frac{3 \pm \sqrt{9 + 64}}{4} \) \( v_s = \frac{3 \pm \sqrt{73}}{4} \)Calculate the value:
\( v_s \approx \frac{3 \pm 8.544}{4} \)We get two possible values for \(v_s\):
- \(v_s \approx \frac{3 + 8.544}{4} = \frac{11.544}{4} \approx 2.886\) km/hr
- \(v_s \approx \frac{3 - 8.544}{4} = \frac{-5.544}{4} \approx -1.386\) km/hr
Since speed must be positive, we choose the positive value.
Final Answer Determination
The calculated speed of the stream is approximately 2.886 km/hr. Rounding to two decimal places gives 2.89 km/hr.
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