Boat Stream Speed Calculation

This problem involves calculating the initial speed of a stream affecting boat travel times downstream and upstream, using given speeds, distances, and time differences.

Stream Speeds Analysis

Identify the given information:

• Boat speed in still water, $v_b = 7$ km/h.
• Distance for both downstream and upstream journeys, $d = 40$ km.
• Let the initial speed of the stream be $v_s$ km/h.
• The stream speed is later considered to be $v'_s = v_s + 2$ km/h.
• The total travel time difference between the two scenarios is 560 minutes.

Convert the time difference from minutes to hours:

Time difference $= \frac{560 \text{ minutes}}{60 \text{ minutes/hour}} = \frac{56}{6} = \frac{28}{3}$ hours.

Time Difference Equation

Scenario 1: Initial Stream Speed ($v_s$)

• Downstream speed $= v_b + v_s = 7 + v_s$ km/h.
• Upstream speed $= v_b - v_s = 7 - v_s$ km/h.
• Time taken $T_1 = (\text{Time downstream}) + (\text{Time upstream})$
• $T_1 = \frac{d}{7 + v_s} + \frac{d}{7 - v_s} = \frac{40}{7 + v_s} + \frac{40}{7 - v_s}$
• Simplify $T_1$: $T_1 = 40 \left( \frac{(7 - v_s) + (7 + v_s)}{(7 + v_s)(7 - v_s)} \right) = 40 \left( \frac{14}{49 - v_s^2} \right) = \frac{560}{49 - v_s^2}$ hours.

Scenario 2: Increased Stream Speed ($v'_s = v_s + 2$)

• Downstream speed $= v_b + v'_s = 7 + (v_s + 2) = 9 + v_s$ km/h.
• Upstream speed $= v_b - v'_s = 7 - (v_s + 2) = 5 - v_s$ km/h.
• Time taken $T_2 = (\text{Time downstream}) + (\text{Time upstream})$
• $T_2 = \frac{d}{9 + v_s} + \frac{d}{5 - v_s} = \frac{40}{9 + v_s} + \frac{40}{5 - v_s}$
• Simplify $T_2$: $T_2 = 40 \left( \frac{(5 - v_s) + (9 + v_s)}{(9 + v_s)(5 - v_s)} \right) = 40 \left( \frac{14}{45 - 4v_s - v_s^2} \right) = \frac{560}{45 - 4v_s - v_s^2}$ hours.

The problem states that the journey took 560 minutes *less* time in Scenario 1 compared to Scenario 2. This means $T_2 - T_1 = \frac{28}{3}$ hours.

Set up the equation:

$ \frac{560}{45 - 4v_s - v_s^2} - \frac{560}{49 - v_s^2} = \frac{28}{3} $

To simplify, divide the equation by 28:

$ \frac{20}{45 - 4v_s - v_s^2} - \frac{20}{49 - v_s^2} = \frac{1}{3} $

Test Initial Speed Option

Direct algebraic solution is complex. We can test the provided options. Option D suggests the initial stream speed is 3 km/h.

Let's verify $v_s = 3$ km/h:

Calculate $T_1$ with $v_s = 3$:

$T_1 = \frac{560}{49 - (3)^2} = \frac{560}{49 - 9} = \frac{560}{40} = 14$ hours.

Calculate $T_2$ with the increased stream speed $v'_s = 3 + 2 = 5$ km/h:

The denominator term is $45 - 4v_s - v_s^2$. Substitute $v_s = 3$:

$45 - 4(3) - (3)^2 = 45 - 12 - 9 = 45 - 21 = 24$.

$T_2 = \frac{560}{24} = \frac{70}{3}$ hours.

Check the time difference:

$T_2 - T_1 = \frac{70}{3} - 14 = \frac{70}{3} - \frac{42}{3} = \frac{28}{3}$ hours.

This calculated difference matches the given time difference of 560 minutes.

Final Answer Verification

The initial speed of the stream is confirmed to be 3 km/h as it satisfies the conditions stated in the problem.