Understanding Motorboat and Current Problems

This problem involves calculating the time taken by a motorboat to travel a certain distance against the current, given the time taken to travel the same distance along the current and the ratio of the motorboat's speed to the current's speed.

When a motorboat travels in water, its speed is affected by the speed of the water current. There are two scenarios:

• Downstream: The boat travels in the same direction as the current. The effective speed is the sum of the motorboat's speed in still water and the speed of the current.
• Upstream: The boat travels against the direction of the current. The effective speed is the difference between the motorboat's speed in still water and the speed of the current.

Setting up the Speeds

We are given the ratio of the speed of the motorboat in still water ($S_m$) to the speed of the current ($S_c$) is 31 : 6.

Let the speed of the motorboat in still water be $31x$ units per hour, and the speed of the current be $6x$ units per hour, where $x$ is a constant.

Based on these speeds, we can find the downstream and upstream speeds:

• Speed Downstream ($S_d$) = Speed of motorboat + Speed of current
• $S_d = S_m + S_c = 31x + 6x = 37x$ units per hour.
• Speed Upstream ($S_u$) = Speed of motorboat - Speed of current
• $S_u = S_m - S_c = 31x - 6x = 25x$ units per hour.

Calculating the Distance Traveled Downstream

The motorboat covers a certain distance along the current (downstream) in 4 hours 10 minutes.

First, let's convert the time into hours:

4 hours 10 minutes = 4 hours $+ \frac{10}{60}$ hours = 4 hours $+ \frac{1}{6}$ hours = $\frac{24+1}{6}$ hours = $\frac{25}{6}$ hours.

Let the distance covered be $D$. The formula for distance is Speed $\times$ Time.

Distance $D = S_d \times T_d$

Where $S_d = 37x$ and $T_d = \frac{25}{6}$ hours.

$D = (37x) \times \left(\frac{25}{6}\right)$ units.

Finding the Time Taken for the Return Journey (Upstream)

The motorboat needs to come back to its initial point, which means it must cover the same distance $D$ against the current (upstream).

The time taken for the return journey (upstream time, $T_u$) is given by the formula:

$T_u = \frac{\text{Distance}}{\text{Speed Upstream}}$

$T_u = \frac{D}{S_u}$

Substitute the values of $D$ and $S_u$:

$T_u = \frac{(37x) \times \left(\frac{25}{6}\right)}{25x}$

We can cancel out $x$ from the numerator and the denominator (since $x$ is a constant representing the speed factor, it cannot be zero):

$T_u = \frac{37 \times \frac{25}{6}}{25}$

$T_u = \frac{37 \times 25}{6 \times 25}$

Cancel out 25 from the numerator and denominator:

$T_u = \frac{37}{6}$ hours.

Converting Upstream Time back to Hours and Minutes

The upstream time is $\frac{37}{6}$ hours.

We can split this into a whole number of hours and a fraction of an hour:

$\frac{37}{6}$ hours = $\frac{36 + 1}{6}$ hours = $\frac{36}{6}$ hours $+ \frac{1}{6}$ hours = 6 hours $+ \frac{1}{6}$ hours.

Now, convert the fraction of an hour to minutes:

$\frac{1}{6}$ hours = $\frac{1}{6} \times 60$ minutes = 10 minutes.

So, the total time taken for the return journey (upstream) is 6 hours 10 minutes.

Summary of Calculations

The time taken by the motorboat to come back to its initial point is 6 hours 10 minutes.

Revision Table: Speed, Time, and Distance Concepts

Additional Information on Boat and Stream Problems

Problems involving boats and streams are common applications of relative speed. Key aspects to remember include:

• The speed of the boat is usually given as its speed in still water.
• The speed of the current is constant for a given problem.
• The direction of travel relative to the current is crucial (downstream or upstream).
• Units must be consistent (e.g., speed in km/h, distance in km, time in hours). Convert units like minutes to hours when necessary for calculations involving hours.
• Distance covered downstream is often equal to the distance covered upstream when the boat returns to its starting point.

Understanding the difference between speed in still water and the effective speed in flowing water is key to solving these problems accurately.