Quadratic Equations: Upstream and Downstream Boat Problems

This lesson solves three motorboat problems on quadratic equations, where a boat moves against the current (upstream) and with the current (downstream). The key is to translate each time condition into an equation and solve for the speed of the stream.

The upstream and downstream rules

Let the speed of the boat in still water be $b$ and the speed of the stream be $x$. Then:

$\text{upstream speed} = b - x \qquad \text{(still minus stream)}$

$\text{downstream speed} = b + x \qquad \text{(still plus stream)}$

For every leg of a journey we use $\text{time} = \dfrac{\text{distance}}{\text{speed}}$.

Problem 1: one hour more upstream

A motorboat whose speed is $18$ km/h in still water takes one hour more to go $24$ km upstream than to return downstream to the same spot. Find the speed of the stream.

Let the speed of the stream be $x$ km/h. Then the upstream speed is $18 - x$ and the downstream speed is $18 + x$, with distance $24$ km each way.

$\text{time upstream} = \frac{24}{18 - x}, \qquad \text{time downstream} = \frac{24}{18 + x}$

The upstream trip takes one hour more:

$\frac{24}{18 - x} - \frac{24}{18 + x} = 1$

Combining the fractions. Take $24$ out and use a common denominator:

$24 \cdot \frac{(18 + x) - (18 - x)}{(18 - x)(18 + x)} = 1$

$24 \cdot \frac{2x}{324 - x^2} = 1 \implies 48x = 324 - x^2$

$x^2 + 48x - 324 = 0$

Solving by the quadratic formula. Here $a = 1$, $b = 48$, $c = -324$:

$b^2 - 4ac = 48^2 - 4(1)(-324) = 2304 + 1296 = 3600, \qquad \sqrt{3600} = 60$

$x = \frac{-48 \pm 60}{2} = 6 \ \text{ or } \ -54$

A speed cannot be negative, so $x = 6$. The speed of the stream is $6$ km/h.

Problem 2: round trip in 4 hours 30 minutes

A motorboat whose speed is $15$ km/h in still water goes $30$ km downstream and returns to the starting point in a total time of $4$ hours $30$ minutes. Find the speed of the stream.

Let the speed of the stream be $x$ km/h. The upstream speed is $15 - x$ and the downstream speed is $15 + x$, with distance $30$ km each way. First convert the total time to hours: $4$ hours $30$ minutes $= 4 + \tfrac{30}{60} = \tfrac{9}{2}$ hours.

$\frac{30}{15 - x} + \frac{30}{15 + x} = \frac{9}{2}$

Combining the fractions.

$30 \cdot \frac{(15 + x) + (15 - x)}{(15 - x)(15 + x)} = \frac{9}{2}$

$30 \cdot \frac{30}{225 - x^2} = \frac{9}{2} \implies \frac{900}{225 - x^2} = \frac{9}{2}$

Cross multiply: $1800 = 9\,(225 - x^2)$, so $200 = 225 - x^2$.

$x^2 = 25 \implies x = \pm 5$

The negative value is rejected, so $x = 5$. The speed of the stream is $5$ km/h.

Problem 3: different distances up and down

The speed of a boat in still water is $8$ km/h. It can go $15$ km upstream and $22$ km downstream in a total of $5$ hours. Find the speed of the stream.

Let the speed of the stream be $x$ km/h. The upstream speed is $8 - x$ and the downstream speed is $8 + x$. Here the two distances differ:

$\frac{15}{8 - x} + \frac{22}{8 + x} = 5$

Combining the fractions.

$\frac{15(8 + x) + 22(8 - x)}{(8 - x)(8 + x)} = 5$

$\frac{120 + 15x + 176 - 22x}{64 - x^2} = 5 \implies \frac{296 - 7x}{64 - x^2} = 5$

Cross multiply: $296 - 7x = 5\,(64 - x^2) = 320 - 5x^2$. Bring every term to one side:

$5x^2 - 7x - 24 = 0$

Solving by factorising. We need two numbers with product $5 \times (-24) = -120$ and sum $-7$; these are $-15$ and $8$:

$5x^2 - 15x + 8x - 24 = 0$

$5x(x - 3) + 8(x - 3) = 0 \implies (x - 3)(5x + 8) = 0$

So $x = 3$ or $x = -\tfrac{8}{5}$. The negative value is rejected, so $x = 3$. The speed of the stream is $3$ km/h.

Key takeaways

• Upstream speed is still water speed minus stream speed; downstream speed is still water speed plus stream speed.
• Write each travel time as distance over speed, then turn the given time condition into a single equation that simplifies to a quadratic.
• Solve the quadratic by the formula or by factorising, and reject the negative root so the speed of the stream is positive.