Understanding Boat and Stream Speed Problems

This problem involves the concept of boat speed in still water and the speed of the stream. When a boat moves upstream, its speed is reduced by the speed of the stream. When it moves downstream, its speed is increased by the speed of the stream. The key relationship is: Time = Distance / Speed.

Let:

• Speed of the boat in still water = \(b\) km/min
• Speed of the stream = \(s\) km/min

Then:

• Speed upstream (\(v_u\)) = \(b - s\) km/min
• Speed downstream (\(v_d\)) = \(b + s\) km/min

We are given two scenarios with different distances and total times. We can use these to set up equations.

Setting Up the Equations

The time taken to travel a certain distance upstream is \( \frac{\text{Distance Upstream}}{v_u} \). The time taken to travel a certain distance downstream is \( \frac{\text{Distance Downstream}}{v_d} \). The total time for each trip is the sum of the upstream and downstream times.

Scenario 1: 3.6 km upstream and 5.4 km downstream in 54 minutes.

Time upstream + Time downstream = Total time

\( \frac{3.6}{v_u} + \frac{5.4}{v_d} = 54 \) (Equation 1)

Scenario 2: 5.4 km upstream and 3.6 km downstream in 58.5 minutes.

Time upstream + Time downstream = Total time

\( \frac{5.4}{v_u} + \frac{3.6}{v_d} = 58.5 \) (Equation 2)

Solving the System of Equations

We have a system of two linear equations with \( \frac{1}{v_u} \) and \( \frac{1}{v_d} \) as the variables. Let \( x = \frac{1}{v_u} \) and \( y = \frac{1}{v_d} \). The equations become:

\( 3.6x + 5.4y = 54 \) (Equation 1')

\( 5.4x + 3.6y = 58.5 \) (Equation 2')

To eliminate decimals, we can multiply both equations by 10:

\( 36x + 54y = 540 \) (Equation 1'')

\( 54x + 36y = 585 \) (Equation 2'')

Now we can solve this system. Multiply Equation 1'' by 3 and Equation 2'' by 2 to make the coefficients of \(x\) equal:

\( 3 \times (36x + 54y) = 3 \times 540 \implies 108x + 162y = 1620 \)

\( 2 \times (54x + 36y) = 2 \times 585 \implies 108x + 72y = 1170 \)

Subtract the second new equation from the first:

\( (108x + 162y) - (108x + 72y) = 1620 - 1170 \)

\( 90y = 450 \)

\( y = \frac{450}{90} = 5 \)

Since \( y = \frac{1}{v_d} \), we have \( \frac{1}{v_d} = 5 \). This means \( v_d = \frac{1}{5} \) km/min. This is the downstream speed.

Now substitute the value of \(y=5\) into Equation 1' (or 1''):

\( 3.6x + 5.4(5) = 54 \)

\( 3.6x + 27 = 54 \)

\( 3.6x = 54 - 27 \)

\( 3.6x = 27 \)

\( x = \frac{27}{3.6} = \frac{270}{36} \)

\( x = \frac{135}{18} = \frac{15}{2} = 7.5 \)

Since \( x = \frac{1}{v_u} \), we have \( \frac{1}{v_u} = 7.5 \). This means \( v_u = \frac{1}{7.5} = \frac{10}{75} = \frac{2}{15} \) km/min. This is the upstream speed.

Calculating Time for 10 km Downstream

We need to find the time taken to travel 10 km downstream. We already found the downstream speed \( v_d = \frac{1}{5} \) km/min.

Time = Distance / Speed

Time = \( \frac{10 \text{ km}}{v_d \text{ km/min}} \)

Time = \( \frac{10}{\frac{1}{5}} \) minutes

Time = \( 10 \times 5 \) minutes

Time = \( 50 \) minutes

Summary of Speeds and Time

From our calculations:

• Upstream speed (\(v_u\)) = \( \frac{2}{15} \) km/min
• Downstream speed (\(v_d\)) = \( \frac{1}{5} \) km/min
• Time for 10 km downstream = 50 minutes

The time taken by the boat in going 10 km downstream is 50 minutes.

Revision Table: Boat and Stream Concepts

Additional Information: Solving Linear Equations

The problem required solving a system of two linear equations with two variables. This is a common technique in various mathematical problems, including those involving speed, distance, and time. Methods for solving such systems include substitution, elimination, and matrix methods. In this solution, we used the elimination method by multiplying the equations and subtracting one from the other to eliminate one variable (\(x\)), allowing us to solve for the other variable (\(y\)). Once one variable is found, it's substituted back into one of the original equations to find the value of the other variable.

In this specific problem, the variables were reciprocals of speeds (\( \frac{1}{v_u} \) and \( \frac{1}{v_d} \)). This is a common pattern in problems where times are given for different distances.