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Hey y'all
This is the Question I came across while preparing for IOQM.
Q) The river flows in a speed of 12m/s. A boat which can travel at 37 m/s speed has to travel from one side of the river to the other (perpendicular distance). The width of the river is 1.4 km. (please find the attached image for a clear view).
Hint: To travel a perpendicular distance of a river that flows, The boat should go in an angle opposite to the flow of river (at an angle θ from the point) to reach the exact perpendicular distance.
To find:
1. The Angle θ
2. The distance from the original point ( travelling opp to river at an angle θ) to the point the boat reaches when the water is still.
3. The minimum time taken by the boat to reach the perpendicular distance when speed of the river is 12 m/s

Sagot :

Answer:

1) 18.92°

2) 1480 m or 1.48 km

3) 40 s

Step-by-step explanation:

We can find the answers for a boat which travels at 37 m/s in a river that flows in a speed of 12 m/s by this way:

In order the boat travels in a perpendicular direction, the resultant of speed with respect to the flow's direction has to be equal to 0 (refer to the drawing), therefore the speed of the boat in x direction [tex](v_x)[/tex] equals to -12 m/s. We can find the angle (θ) by using the trigonometric formula:

1)

[tex]\begin{aligned}sin\theta&=\frac{|v_x|}{|v|} \\\\sin\theta&=\frac{12}{37} \\\\\theta&=asin\left(\frac{12}{37} \right)\\\\\theta&\approx\bf 18.92^o\end{aligned}[/tex]

2)

Since the distance in y direction [tex](s_y)[/tex], which is the river's width, is 1.4 km = 1400 m, then we can find the distance [tex](s)[/tex] by using the trigonometric formula:

[tex]\begin{aligned}cos\theta&=\frac{s_y}{s} \\\\cos(18.92^o)&=\frac{1400}{s} \\\\s&=\frac{1400}{cos(18.92)}\\\\s&=\bf 1480\ m\ or\ 1.48\ km\end{aligned}[/tex]

3) To find the time taken by the boat, we can use the linear motion with constant speed formula:

[tex]\boxed{s=vt}[/tex]

Given:

  • [tex]s=1480\ m[/tex]
  • [tex]v=37\ m/s[/tex]

Then:

[tex]\begin{aligned}\\s&=vt\\1480&=37t\\t&=1480\div37\\t&=\bf40\ s\end{aligned}[/tex]

View image Karmenchong
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