Understanding the Motorboat Speed Problem

This question asks us to find the speed of a motorboat in still water given its travel time upstream and downstream in a stream with a known speed. This is a classic boat and stream problem, which involves understanding how the speed of the water affects the boat's speed.

Defining Variables and Knowns

Let's define the speeds involved:

• Let v be the speed of the motorboat in still water (in km/h). This is what we need to find.
• Let s be the speed of the stream (in km/h). We are given that s = 3 km/h.

When the boat travels:

• Upstream: The stream opposes the boat's movement. The effective speed is the boat's speed minus the stream's speed: Speed upstream = v - s. For the boat to move upstream, the boat's speed must be greater than the stream's speed (v > s).
• Downstream: The stream helps the boat's movement. The effective speed is the boat's speed plus the stream's speed: Speed downstream = v + s.

We are also given:

• The distance traveled upstream is 12 km.
• The distance traveled downstream (back to the starting point) is also 12 km.
• The total time taken for the round trip (upstream and downstream) is 60 minutes, which is equal to 1 hour.

Setting Up the Equation using Time, Distance, and Speed

The fundamental relationship between time, distance, and speed is: Time = Distance / Speed.

The total time for the round trip is the sum of the time taken to travel upstream and the time taken to travel downstream.

Total Time = Time Upstream + Time Downstream

Using the formula Time = Distance / Speed:

• Time Upstream = Distance Upstream / Speed Upstream = $\frac{12}{v - s}$
• Time Downstream = Distance Downstream / Speed Downstream = $\frac{12}{v + s}$

Substituting the known values (s = 3 km/h, Total Time = 1 hour):

$\frac{12}{v - 3} + \frac{12}{v + 3} = 1$

Solving for the Speed of the Motorboat in Still Water

Now, we need to solve the equation $\frac{12}{v - 3} + \frac{12}{v + 3} = 1$ for v.

First, find a common denominator on the left side, which is $(v - 3)(v + 3)$:

$\frac{12(v + 3) + 12(v - 3)}{(v - 3)(v + 3)} = 1$

Expand the numerator:

$\frac{12v + 36 + 12v - 36}{(v - 3)(v + 3)} = 1$

Simplify the numerator:

$\frac{24v}{(v - 3)(v + 3)} = 1$

Use the difference of squares formula $(a - b)(a + b) = a^2 - b^2$ for the denominator:

$\frac{24v}{v^2 - 3^2} = 1$

$\frac{24v}{v^2 - 9} = 1$

Multiply both sides by $(v^2 - 9)$ to remove the denominator:

$24v = v^2 - 9$

Rearrange the terms to form a standard quadratic equation $av^2 + bv + c = 0$:

$v^2 - 24v - 9 = 0$

Now, we use the quadratic formula to solve for v:

$v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our equation, $a = 1$, $b = -24$, and $c = -9$. Substitute these values into the formula:

$v = \frac{-(-24) \pm \sqrt{(-24)^2 - 4(1)(-9)}}{2(1)}$

$v = \frac{24 \pm \sqrt{576 + 36}}{2}$

$v = \frac{24 \pm \sqrt{612}}{2}$

Now, simplify the square root $\sqrt{612}$. We can factor 612 to find perfect square factors:

$612 = 4 \times 153 = 4 \times 9 \times 17 = 36 \times 17$

So, $\sqrt{612} = \sqrt{36 \times 17} = \sqrt{36} \times \sqrt{17} = 6\sqrt{17}$

Substitute this back into the expression for v:

$v = \frac{24 \pm 6\sqrt{17}}{2}$

Divide both terms in the numerator by 2:

$v = 12 \pm 3\sqrt{17}$

We have two possible solutions for v: $12 + 3\sqrt{17}$ and $12 - 3\sqrt{17}$.

Remember that the speed of the boat in still water (v) must be greater than the speed of the stream (s = 3 km/h) for the boat to be able to travel upstream. Let's approximate the value of $3\sqrt{17}$. Since $\sqrt{16} = 4$ and $\sqrt{25} = 5$, $\sqrt{17}$ is slightly more than 4. So, $3\sqrt{17}$ is slightly more than $3 \times 4 = 12$.

• For $v = 12 + 3\sqrt{17}$: This value is clearly greater than 12, and thus greater than 3. This is a valid speed.
• For $v = 12 - 3\sqrt{17}$: Since $3\sqrt{17}$ is slightly more than 12, $12 - 3\sqrt{17}$ will be negative. A negative speed is not physically possible in this context. Also, $v$ must be greater than $s=3$. $12 - 3\sqrt{17}$ is less than $12 - 12 = 0$, so it is not greater than 3. This solution is extraneous.

Therefore, the only valid solution for the speed of the motorboat in still water is:

$v = 12 + 3\sqrt{17}$

We can factor out a 3 from this expression:

$v = 3(4 + \sqrt{17})$

Summary of the Calculation Steps

1. Identify knowns: stream speed (s), distance (d), total time (T).
2. Define unknown: boat speed in still water (v).
3. Write expressions for upstream speed (v - s) and downstream speed (v + s).
4. Write expressions for time upstream ($\frac{d}{v-s}$) and time downstream ($\frac{d}{v+s}$).
5. Set up the equation: Time Upstream + Time Downstream = Total Time.
6. Substitute values and solve the resulting equation for v. This leads to a quadratic equation.
7. Use the quadratic formula to find possible values for v.
8. Reject any physically impossible solutions (like negative speed or speed less than stream speed).
9. Simplify the valid solution.

Final Answer for Motorboat Speed

The speed of the motorboat in still water is $3(4 + \sqrt{17})$ km/h.

Revision Table: Boat and Stream Concepts

Additional Information: Solving Quadratic Equations

In many physics and math problems, including those involving speeds, you might encounter quadratic equations in the form $ax^2 + bx + c = 0$. These can be solved using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The term inside the square root, $b^2 - 4ac$, is called the discriminant ($\Delta$).

• If $\Delta > 0$, there are two distinct real solutions.
• If $\Delta = 0$, there is exactly one real solution (a repeated root).
• If $\Delta < 0$, there are no real solutions (solutions are complex numbers).

In our boat and stream problem, the variable was 'v' instead of 'x', and we had $v^2 - 24v - 9 = 0$. The discriminant was $(-24)^2 - 4(1)(-9) = 576 + 36 = 612$, which is positive, giving us two real solutions. We then used the physical constraints of the problem (speed must be positive and greater than stream speed) to choose the correct solution.