Question

boat 'a' is moving with speed of 10 km/h in west direction. in 100 km south of a is boat 'b' moving towards north (speed of boat 'b' is also 10 km/h). in what time will the two boats be at the shortest distance from each other

Answer 1

Question #57161, Physics / Mechanics | Relativity

boat 'a' is moving with speed of $10\mathrm{km/h}$ in west direction. In $100\mathrm{km}$ south of a is boat 'b' moving towards north (speed of boat 'b' is also $10\mathrm{km/h}$). In what time will the two boats be at the shortest distance from each other

Solution:

Distance between two boats:

$$d = \sqrt {x ^ {2} (t) + y ^ {2} (t)}$$

Displacement of boat 1:

$$x (t) = 1 0 t$$

Displacement of boat 2:

$$y (t) = - 1 0 0 + 1 0 t$$

Combining (1), (2) and (3):

$$d = \sqrt {1 0 0 t ^ {2} + (1 0 t - 1 0 0) ^ {2}};$$ $$d = 1 0 \sqrt {2 t ^ {2} - 1 0 t + 1 0 0}$$

To find the smallest distance between the boats, one should minimize function (4).

Considering the interval $0 \leq t \leq 10$, because after 10 hours boat 2 will reach $x$-axis and distance will be $100\mathrm{km}$ (displacement of boat 1 during 10 hours), and will grow further, as seen from the graph.

$$d ^ {\prime} = \frac {2 0 \times (2 t - 5)}{\sqrt {2 t ^ {2} - 1 0 t + 1 0 0}};$$ $$\begin{array}{l} d' = 0; \\ \frac{20 \times (2t - 5)}{\sqrt{2t^2 - 10t + 100}} = 0; \\ t = 2.5; \, d(2.5) = 93.54 \, \text{km} \end{array}$$

Answer: in 2.5 hours after initial moment.

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