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A river $200 \mathrm{~m}$ wide is flowing at a rate of $3.0 \mathrm{~m} / \mathrm{s}$. A boat is sailing at a velocity of $15 \mathrm{~m} / \mathrm{s}$ with respect to the water in a direction perpendicular to the river. How far from the point directly opposite to the starting point does the boat reach on the opposite bank?
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Verified Answer
The correct answer is:
$40 \mathrm{~m}$
The given situation is shown in the following figure
Velocity of river, $v_r=3 \mathrm{~ms}^{-1}$
Velocity of boat, $v_b=15 \mathrm{~ms}^{-1}$
Width of river $=A B=200 \mathrm{~m}$
According to question, we have to calculate distance $B C$.
Time taken by boat to cross the river,
$t=\frac{A B}{v_b}=\frac{200}{15}=\frac{40}{3} \mathrm{~s}$
Since, boat covers horizontal distance $B C$ due to it own velocity.
Hence, $\quad B C=v_r \times t=3 \times \frac{40}{3}=40 \mathrm{~m}$
Velocity of river, $v_r=3 \mathrm{~ms}^{-1}$
Velocity of boat, $v_b=15 \mathrm{~ms}^{-1}$
Width of river $=A B=200 \mathrm{~m}$
According to question, we have to calculate distance $B C$.
Time taken by boat to cross the river,
$t=\frac{A B}{v_b}=\frac{200}{15}=\frac{40}{3} \mathrm{~s}$
Since, boat covers horizontal distance $B C$ due to it own velocity.
Hence, $\quad B C=v_r \times t=3 \times \frac{40}{3}=40 \mathrm{~m}$
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