The boat left point A for point B, the distance between points is 210 km. The boat spent 4 hours less on the return trip(from point B to point A) than on the way there(To point B). The speed of the river is 3 km/h. Find the boat's own speed.

Let the boat's own speed in still water be \( v \) km/h.

### Problem breakdown:
- The distance between point A and point B is 210 km.
- The boat spent 4 hours less on the return trip.
- The speed of the river (current) is 3 km/h.

### Speeds with respect to the current:
- On the way to point B (upstream), the effective speed of the boat is \( v - 3 \) km/h because the current is slowing the boat down.
- On the return trip to point A (downstream), the effective speed of the boat is \( v + 3 \) km/h because the current is aiding the boat.

### Time equations:
- The time taken to travel from point A to point B (upstream) is \( \frac{210}{v - 3} \) hours.
- The time taken to return from point B to point A (downstream) is \( \frac{210}{v + 3} \) hours.

Since the return trip took 4 hours less than the trip to point B, we can set up the following equation:

\[
\frac{210}{v - 3} - \frac{210}{v + 3} = 4
\]

### Solving the equation:

1. First, factor out the 210:
\[
210 \left( \frac{1}{v - 3} - \frac{1}{v + 3} \right) = 4
\]

2. Use the common denominator to combine the terms inside the parentheses:
\[
210 \left( \frac{(v + 3) - (v - 3)}{(v - 3)(v + 3)} \right) = 4
\]
\[
210 \left( \frac{6}{v^2 - 9} \right) = 4
\]

3. Simplify and solve for \( v^2 \):
\[
\frac{1260}{v^2 - 9} = 4
\]
\[
1260 = 4(v^2 - 9)
\]
\[
1260 = 4v^2 - 36
\]
\[
1296 = 4v^2
\]
\[
v^2 = \frac{1296}{4} = 324
\]
\[
v = \sqrt{324} = 18
\]

### Conclusion:
The boat's own speed in still water is 18 km/h.

Do you want more details or have any questions?

Here are 5 related questions:
1. What would happen if the river's speed increased to 5 km/h?
2. How long would each trip take at the boat's own speed of 18 km/h?
3. How does the boat's speed in still water affect travel times upstream and downstream?
4. Can this method be applied to other types of vehicles moving in currents, such as planes with wind resistance?
5. How would you solve this problem if the trip distance were doubled?

**Tip:** When dealing with motion problems, pay attention to the relative speed when currents or winds are involved.

The boat left point A for point B, the distance between points is 210 km. The boat spent 4 hours less on the return trip (from point B to point A) than on the way there (To point B). The speed of the river is 3 km/h. Find the boat's own speed.

Learn how to solve a boat travel problem involving relative speed, river current, and time difference. The problem involves a boat traveling 210 km between two points with a river speed of 3 km/h, requiring algebraic equations and the quadratic formula. Suitable for students in Grades 9-11.

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