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Problem of the Week
Problem C and Solution
See You No More

Problem

Two boats are travelling away from each other in opposite directions. One boat is travelling east at the constant speed of \(8\) km/h and the other boat is travelling west at a different constant speed.

At one point, the boat travelling east was \(200\) m east of the boat travelling west, but \(15\) minutes later they lose sight of each other.

If the visibility at sea that day was 5 km, determine the constant speed of the boat travelling west.

Solution

We will call the boat travelling east Boat \(A\), and the boat travelling west Boat \(B\).

Boat \(A\) is travelling at a constant speed of \(8\) km/h.

Using the formula, distance \(=\) speed \(\times\) time, in \(15\) minutes Boat \(A\) will travel \(8\ \dfrac{\text{km}}{\text{h}} \times \dfrac{15}{60}\ \text{h} = 2\) km.

The visibility at sea is \(5\) km. Thus, Boat \(A\) and Boat \(B\) will be in sight of one another until they are \(5\) km apart. We are given that Boat \(A\) and Boat \(B\) are in sight of one another for \(15\) minutes. Thus, after \(15\) minutes Boat \(A\) and Boat \(B\) must be \(5\) km apart.

Since Boat \(A\) and Boat \(B\) start out \(200 \mbox{ m} = 0.2 \mbox{ km}\) apart and Boat \(A\) travels \(2\) km in \(15\) minutes, Boat \(B\) must travel \(5 - 0.2 - 2 = 2.8\) km in \(15\) minutes.

Since Boat \(B\) travelled \(2.8\) km in \(15\) minutes, using the formula
speed \(=\) distance \(\div\) time, Boat \(B\) must have been travelling at a speed of \(2.8 \ \text{km} \div \dfrac{15}{60} \ \text{h} = 2.8 \times \dfrac{60}{15} = 11.2\) km/h.

Therefore, Boat \(B\) was travelling at a speed of \(11.2\) km/h.