Two boat landings are 1.0 km apart on the same bank of a stream that flows at 1.0 km/h. A motorboat makes the round trip between the two landings in 50 minutes. What is the speed of the boat relative to the water?

Refer to the figure shown below.

Let V = speed of the boat relative to the water

Given:

u = 1 km/h the speed of flowing water.

When traveling downstream from A to B, the actual speed of the boat is

V₁ = V + u = V + 1 km/h

When traveling upstream from B to A, the actual speed of the boat is

V₂ = V - u = V - 1 km/h

Because the distance Ab is 1 km, the time taken for the round trip is

t = (1 km) / (V+1 km/h) + (1 km) / (v-1 km/h)

= (V-1 + V+1) / (V² - 1)

= (2V) / (V² - 1)

The time for the round trip is 50 min = 5/6 h.

Therefore

(2V) / (V² - 1) = 5/6

5 (V² - 1) = 12V

5V² - 12V - 5 = 0

Solve with the quadratic formula.

V = (1/10) * [12 + / - √ (144 + 100) ] = 2.762 or - 0.362 km/h

Ignore negative speed, so that

V = 2.762 k/h

Answer:

The speed of the boat relative to the water is 2.76 km/h (nearest hundredth)