Question 21: Motion in 2D & Vectors

Question

A river is flowing from west to east direction with speed of $9 \mathrm{~km} \mathrm{~h}^{-1}$. If a boat capable of moving at a maximum speed of $27 \mathrm{~km} \mathrm{~h}^{-1}$ in still water, crosses the river in half a minute, while moving with maximum speed at an angle of $150^{\circ}$ to direction of river flow, then the width of the river is :

Accepted Answer

Step 1: Find the boatÃ¢ÂÂs speed across the river
The boat's maximum speed is $27~\mathrm{km/h}$, and it makes an angle of $150^\circ$ with the direction of the river.
The part of the boatÃ¢ÂÂs speed that helps it cross the river straight (perpendicular to the river) is found by: $27 	imes \cos 60^\circ$.
Since $\cos 60^\circ = \frac{1}{2}$, this gives: $27 	imes \frac{1}{2} = \frac{27}{2}~\mathrm{km/h}$.
Step 2: Change speed to meters per second
$\frac{27}{2}~\mathrm{km/h} = \frac{27}{2} 	imes \frac{1000}{3600} = \frac{27}{2} 	imes \frac{5}{18}~\mathrm{m/s}$.
Step 3: Use the time the boat takes
The boat takes $30$ seconds to cross.
Distance = speed ÃÂ time
So the width of the river is:
$$\mathrm{S} = \frac{27}{2} 	imes \frac{5}{18} 	imes 30~\mathrm{m}$$
Step 4: Do the calculation
Calculate the value:
$\frac{27}{2} 	imes \frac{5}{18} 	imes 30 = 112.5~\mathrm{m}$
So, the width of the river is $112.5$ meters.

Answer

112.5 m

Answer

75 m

Answer

300 m

Answer

$112.5 	imes \sqrt{3} \mathrm{~m}$

Motion in 2D & Vectors

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