Calculating Speedboat Stream Speed

The problem asks for the speed of the stream (\(v_b\)) given the speedboat's speed in still water (\(v_s\)), the distance traveled downstream, and the total round trip time.

Given:

• Speed of speedboat in still water, \(v_s = 16\) km/hr
• Distance downstream, \(d = 40\) km
• Total time, \(T = 6\) hours 24 minutes

Step 1: Convert Total Time to Hours

Convert the total time to hours using fractions for precision:

Step 2: Define Downstream and Upstream Speeds

Let the speed of the stream be \(v_b\) km/hr.

• Speed downstream = Speed of boat + Speed of stream = \(v_s + v_b = 16 + v_b\) km/hr
• Speed upstream = Speed of boat - Speed of stream = \(v_s - v_b = 16 - v_b\) km/hr

Step 3: Formulate the Time Equation

The time taken for each leg of the journey is calculated as Distance / Speed.

• Time downstream (\(t_{down}\)) = \(\frac{40}{16 + v_b}\)
• Time upstream (\(t_{up}\)) = \(\frac{40}{16 - v_b}\)
• Total time \(T = t_{down} + t_{up}\)

Step 4: Solve for the Speed of the Stream (\(v_b\))

Simplify the equation:

1. Divide the entire equation by 40: \( \frac{1}{16 + v_b} + \frac{1}{16 - v_b} = \frac{32}{5 \times 40} = \frac{32}{200} = \frac{4}{25} \)
2. Combine the fractions on the left side using a common denominator \((16 + v_b)(16 - v_b)\): \( \frac{(16 - v_b) + (16 + v_b)}{(16 + v_b)(16 - v_b)} = \frac{4}{25} \) \( \frac{32}{16^2 - v_b^2} = \frac{4}{25} \) \( \frac{32}{256 - v_b^2} = \frac{4}{25} \)
3. Cross-multiply: \( 32 \times 25 = 4 \times (256 - v_b^2) \) \( 800 = 1024 - 4v_b^2 \)
4. Rearrange to solve for \(v_b^2\): \( 4v_b^2 = 1024 - 800 \) \( 4v_b^2 = 224 \) \( v_b^2 = \frac{224}{4} \) \( v_b^2 = 56 \)
5. Take the square root to find \(v_b\): \( v_b = \sqrt{56} \) \( v_b = \sqrt{4 \times 14} \) \( v_b = 2\sqrt{14} \)

The speed of the stream is \(2\sqrt{14}\) km/hr.