
The width of a river is 1km. The velocity of the boat is $5km/hr$ . The boat covered the width of the river with the shortest will possible in 15 min. Then the velocity of river stream is:
$\begin{align}
& \text{A}\text{. 3km/hr} \\
& \text{B}\text{. 4km/hr} \\
& \text{C}\text{. }\sqrt{20}km/hr \\
& \text{D}\text{. }\sqrt{41}km/hr \\
\end{align}$
Answer
572.4k+ views
Hint: To solve this question we will use the concept of vector subtraction. Since, the velocity of the boat is making an angle with the velocity of the river stream, the velocity in the perpendicular direction between the banks of the river is given by the vector subtraction of the velocities. Then, put this value on the law of motion to find the required velocity.
Complete step-by-step answer:
Given in the question that, the width of the river is, $d=1km$ .
The velocity of the boat is $u=5km/hr$
Time required by the boat to cross the river with the shortest possible path is $t=15\min =\dfrac{15}{60}hr=\dfrac{1}{4}hr$
We need to find the velocity of the river stream.
Now, the shortest possible path will be the perpendicular distance between the banks of the river as shown in the figure.
For the resultant velocity to be in the perpendicular direction, the velocity of the boat will make an angle with the velocity of the river stream as shown in figure.
Let the velocity of the river stream is v.
So, the resultant velocity of the boat can be given as,
$V=\sqrt{{{u}^{2}}-{{v}^{2}}}$
So, the time required to cross the distance d can be given as,
$t=\dfrac{d}{V}$
Putting the required values in the above equation, we get that,
$\begin{align}
& t=\dfrac{1}{\sqrt{{{5}^{2}}-{{v}^{2}}}} \\
& \dfrac{1}{4}=\dfrac{1}{\sqrt{{{5}^{2}}-{{v}^{2}}}} \\
& \dfrac{1}{16}=\dfrac{1}{25-{{v}^{2}}} \\
& 25-{{v}^{2}}=16 \\
& {{v}^{2}}=9 \\
& v=3km/hr \\
\end{align}$
So, to cross the river with the shortest possible distance in 15 min with velocity of the boat $5km/hr$ , the velocity of the river stream must be $3km/hr$ .
So, the correct answer is “Option A”.
Note: We can directly say that to cross the river to the point directly opposite to the bank, the velocity of the boat should always be larger than the velocity of flow of water of the river. If the velocity of the flow of the river is larger than the velocity of the boat, he can never reach the point directly opposite to the bank.
Complete step-by-step answer:
Given in the question that, the width of the river is, $d=1km$ .
The velocity of the boat is $u=5km/hr$
Time required by the boat to cross the river with the shortest possible path is $t=15\min =\dfrac{15}{60}hr=\dfrac{1}{4}hr$
We need to find the velocity of the river stream.
Now, the shortest possible path will be the perpendicular distance between the banks of the river as shown in the figure.
For the resultant velocity to be in the perpendicular direction, the velocity of the boat will make an angle with the velocity of the river stream as shown in figure.
Let the velocity of the river stream is v.
So, the resultant velocity of the boat can be given as,
$V=\sqrt{{{u}^{2}}-{{v}^{2}}}$
So, the time required to cross the distance d can be given as,
$t=\dfrac{d}{V}$
Putting the required values in the above equation, we get that,
$\begin{align}
& t=\dfrac{1}{\sqrt{{{5}^{2}}-{{v}^{2}}}} \\
& \dfrac{1}{4}=\dfrac{1}{\sqrt{{{5}^{2}}-{{v}^{2}}}} \\
& \dfrac{1}{16}=\dfrac{1}{25-{{v}^{2}}} \\
& 25-{{v}^{2}}=16 \\
& {{v}^{2}}=9 \\
& v=3km/hr \\
\end{align}$
So, to cross the river with the shortest possible distance in 15 min with velocity of the boat $5km/hr$ , the velocity of the river stream must be $3km/hr$ .
So, the correct answer is “Option A”.
Note: We can directly say that to cross the river to the point directly opposite to the bank, the velocity of the boat should always be larger than the velocity of flow of water of the river. If the velocity of the flow of the river is larger than the velocity of the boat, he can never reach the point directly opposite to the bank.
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