Analyzing Boat Speed and Water Current

This problem involves understanding the motion of a boat in water, specifically how its speed is affected by the water current when moving upstream and downstream. We are given the time taken to cover a certain distance in both directions and need to find the ratio of the boat's speed in still water to the speed of the water current.

Defining Speeds and Distances

Let's define the key speeds involved:

• Let \(v_b\) be the speed of the boat in still water.
• Let \(v_c\) be the speed of the water current.

When the boat moves upstream, it is going against the water current, so the effective speed is reduced. When the boat moves downstream, it is going with the water current, so the effective speed is increased.

• Speed upstream = \(v_b - v_c\)
• Speed downstream = \(v_b + v_c\)

Let \(D\) be the distance covered in one direction (from starting point to the destination and back).

Converting Time Units

The time taken for the upstream journey is given as 8 hours 48 minutes. We need to convert this time entirely into hours.

• 8 hours 48 minutes = 8 hours + 48 minutes
• To convert minutes to hours, divide by 60: \(48 \text{ minutes} = \frac{48}{60} \text{ hours} = \frac{4}{5} \text{ hours} = 0.8 \text{ hours}\).
• So, the upstream time is \(8 + 0.8 = 8.8\) hours.

The time taken for the downstream journey is given as 4 hours.

Setting Up Equations with Distance, Speed, and Time

We know the formula: Distance = Speed \(\times\) Time.

Using this formula, we can set up equations for both the upstream and downstream journeys:

• For the upstream journey: \(D = (v_b - v_c) \times 8.8\)
• For the downstream journey: \(D = (v_b + v_c) \times 4\)

Since the distance \(D\) is the same for both journeys, we can equate the right-hand sides of the two equations:

\((v_b - v_c) \times 8.8 = (v_b + v_c) \times 4\)

Solving for the Ratio of Speeds

Now, we need to solve this equation to find the ratio \(v_b : v_c\).

First, distribute the numbers on both sides of the equation:

\(8.8 v_b - 8.8 v_c = 4 v_b + 4 v_c\)

Next, gather the terms involving \(v_b\) on one side and the terms involving \(v_c\) on the other side. Let's move \(4 v_b\) to the left side and \(-8.8 v_c\) to the right side.

\(8.8 v_b - 4 v_b = 4 v_c + 8.8 v_c\)

Perform the subtraction on the left and the addition on the right:

\(4.8 v_b = 12.8 v_c\)

To find the ratio \(v_b : v_c\), we can divide both sides by \(v_c\) and by 4.8:

\(\frac{v_b}{v_c} = \frac{12.8}{4.8}\)

To simplify the ratio \(\frac{12.8}{4.8}\), we can remove the decimal points by multiplying the numerator and the denominator by 10:

\(\frac{v_b}{v_c} = \frac{128}{48}\)

Now, simplify the fraction \(\frac{128}{48}\) by dividing the numerator and denominator by their greatest common divisor. We can see that both numbers are divisible by 16 (128 = 16 \(\times\) 8, 48 = 16 \(\times\) 3).

\(\frac{v_b}{v_c} = \frac{128 \div 16}{48 \div 16} = \frac{8}{3}\)

So, the ratio of the speed of the boat in still water (\(v_b\)) to the speed of the water current (\(v_c\)) is 8 : 3.

Conclusion on Boat Speed Ratio

The calculated ratio of the speed of the boat in still water to that of the water current is 8 : 3. This means that for every 8 units of speed of the boat in still water, the speed of the current is 3 units.

Revision Table: Boat and Stream Concepts

Additional Information: Solving Boat and Stream Problems

Boat and stream problems are common in time, speed, and distance calculations. They typically involve finding the speed of the boat, the speed of the current, the distance, or the time taken for journeys upstream or downstream. The core idea is to correctly calculate the effective speed by adding or subtracting the speed of the current from the boat's speed in still water.

Key points to remember:

• Always ensure that units of speed, time, and distance are consistent. Convert minutes to hours or vice versa if necessary.
• The distance covered upstream and downstream is often the same in return journey problems, allowing you to equate the distance formulas.
• If you know the upstream speed (\(S_u\)) and downstream speed (\(S_d\)), you can directly find the speed of the boat and current:
• Speed of boat in still water \(v_b = \frac{S_d + S_u}{2}\)
• Speed of current \(v_c = \frac{S_d - S_u}{2}\)

In this problem, we found the ratio by solving the equation \(8.8(v_b - v_c) = 4(v_b + v_c)\). We could also have found the speeds relative to each other. Let \(S_u = v_b - v_c\) and \(S_d = v_b + v_c\). We have \(D = S_u \times 8.8\) and \(D = S_d \times 4\). So, \(S_u \times 8.8 = S_d \times 4\), which means \(\frac{S_d}{S_u} = \frac{8.8}{4} = \frac{88}{40} = \frac{11}{5}\). \(S_d : S_u = 11 : 5\). So, \(v_b + v_c\) is proportional to 11, and \(v_b - v_c\) is proportional to 5. Using the formulas: \(v_b \propto \frac{11+5}{2} = \frac{16}{2} = 8\) \(v_c \propto \frac{11-5}{2} = \frac{6}{2} = 3\) Thus, \(v_b : v_c = 8 : 3\). This confirms our previous calculation.