A man wishes to cross a river of width $120 m$ by a motorboat. His rowing speed in still water is $3 m / s$ and his maximum walking speed is $1 m / s$ . The river flows with velocity of $4 m / s$ . The minimum time which he takes to reach his destination in seconds is 20x. Find the value of x.

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Solution

Let us consider the angle between the direction of river flow and the man to be $θ$ and we can resolve the velocity horizontally and vertically.
Time to cross the river
(I) $t_{1} = \frac{120}{3 c o s θ} = \frac{40}{cos θ} = 40 sec θ$
Drift along the river $x = (4 - 3 sin θ) (\frac{40}{cos θ})$
$= (160 sec θ - 120 tan θ)$
To reach directly opposite, this drift will be covered by walking speed.
Time taken in this,
(II) $t_{2} = \frac{160 sec θ - 120 tan θ}{1} = 160 sec θ - 120 tan θ$
$∴$ Total time taken
$t = t_{1} + t_{2} = (200 sec θ - 120 tan θ)$
For t to be minimum, $\frac{d t}{d θ} = 0$
or $200 sec θ tan θ - 120 {sec}^{2} θ = 0$
or $θ = {sin}^{- 1} (\frac{3}{5})$
$t_{m i n} = 200 sec θ - 120 tan θ$ (where, $sin θ = \frac{3}{5})$
$= 200 \times \frac{5}{4} - 120 \times \frac{3}{4}$
$= 250 - 90 = 160 s = 2 min 40 s = 160 s = 20 \times 8 s$

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A man wishes to cross a river of width 120 m by a motorboat. His rowing speed in still water is 3 m/s and his maximum walking speed is 1 m/s. The river flows with velocity of 4 m/s. The minimum time which he takes to reach his destination in seconds is 20x. Find the value of x.

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