Understanding Boat and Stream Speed Problems

This problem involves the concepts of boat speed in still water and the speed of the water stream. When a boat travels downstream, its speed is the sum of its own speed and the stream's speed. When it travels upstream, the stream opposes the boat's movement, so its speed is the difference between its own speed and the stream's speed.

Defining Variables

• Let \(v_b\) represent the speed of the boat in still water.
• Let \(v_s\) represent the speed of the stream.
• Let \(D\) represent the distance covered in one direction (downstream or upstream).
• Let \(t_d\) represent the time taken to travel downstream.
• Let \(t_u\) represent the time taken to travel upstream.

Key Formulas

• Speed downstream = \(v_b + v_s\)
• Speed upstream = \(v_b - v_s\)
• Distance = Speed × Time

Given Information

From the problem statement, we are given:

• Time taken downstream, \(t_d = 4\) hours.
• Time taken upstream, \(t_u = 6\) hours.
• The distance covered downstream and upstream is the same, \(D\).

Setting up the Equations

Using the formula Distance = Speed × Time, we can write equations for the downstream and upstream journeys:

For the downstream journey:

\[D = (v_b + v_s) \times t_d\]

Substituting the given time:

\[D = (v_b + v_s) \times 4 \quad \cdots (1)\]

For the upstream journey:

\[D = (v_b - v_s) \times t_u\]

Substituting the given time:

\[D = (v_b - v_s) \times 6 \quad \cdots (2)\]

Solving for the Ratio

Since the distance \(D\) is the same for both journeys, we can equate equation (1) and equation (2):

\[(v_b + v_s) \times 4 = (v_b - v_s) \times 6\]

Now, we will solve this equation to find the ratio of \(v_s\) to \(v_b\).

Distribute the numbers on both sides:

\[4v_b + 4v_s = 6v_b - 6v_s\]

Gather terms with \(v_s\) on one side and terms with \(v_b\) on the other side. Add \(6v_s\) to both sides:

\[4v_b + 4v_s + 6v_s = 6v_b - 6v_s + 6v_s\]

\[4v_b + 10v_s = 6v_b\]

Subtract \(4v_b\) from both sides:

\[4v_b + 10v_s - 4v_b = 6v_b - 4v_b\]

\[10v_s = 2v_b\]

We need to find the ratio \(v_s : v_b\), which is \(\frac{v_s}{v_b}\). To get this ratio, divide both sides of the equation by \(10v_b\):

\[\frac{10v_s}{10v_b} = \frac{2v_b}{10v_b}\]

Simplify both sides:

\[\frac{v_s}{v_b} = \frac{2}{10}\]

\[\frac{v_s}{v_b} = \frac{1}{5}\]

So, the ratio of the speed of the stream to the speed of the boat in still water (\(v_s : v_b\)) is \(1 : 5\).

Result

The ratio of the speed of the stream to the speed of the boat in still water is \(1 : 5\).

Revision Table: Boat and Stream Formulas

Additional Information: Relating Speed and Time

In problems where distance is constant, speed and time are inversely proportional. This means if speed increases, time decreases, and vice versa. In this problem, the boat travels slower upstream than downstream because the stream opposes it. Since the distance is the same, the time taken upstream is greater than the time taken downstream, which is consistent with the given values (6 hours upstream > 4 hours downstream).

We can also use the alternative formulas for \(v_b\) and \(v_s\) if we first find the speeds. Let \(v_d\) be the speed downstream and \(v_u\) be the speed upstream. If the distance is \(D\):

• \(v_d = \frac{D}{t_d} = \frac{D}{4}\)
• \(v_u = \frac{D}{t_u} = \frac{D}{6}\)

Using the formulas for \(v_b\) and \(v_s\):

• \(v_b = \frac{v_d + v_u}{2} = \frac{\frac{D}{4} + \frac{D}{6}}{2} = \frac{\frac{3D+2D}{12}}{2} = \frac{5D}{24}\)
• \(v_s = \frac{v_d - v_u}{2} = \frac{\frac{D}{4} - \frac{D}{6}}{2} = \frac{\frac{3D-2D}{12}}{2} = \frac{D}{24}\)

Now, find the ratio \(v_s : v_b\):

\[\frac{v_s}{v_b} = \frac{\frac{D}{24}}{\frac{5D}{24}} = \frac{D}{24} \times \frac{24}{5D} = \frac{1}{5}\]

This confirms the result obtained using the first method. Both approaches lead to the same ratio of stream speed to boat speed in still water.