Question

A swimmer swims in still water at a speed $=5km/hr$. He enters a $200m$ wide river, having river flow speed $=4km/hr$ at point A and proceeds to swim at an angle of ${127}^{\circ }$ with the river flow direction. Another point $B$ is located directly across $A$ on the other side. The swimmer lands on the other bank at a point $C$, from which he walks the distance $CB$ with a speed $=3km/hr$. The total time in which he reaches from $A$ to $B$ is

• A. $5minutes$
• B. $4minutes$
• C. $3minutes$
• D. None

Answer 1

The correct option is B $4minutes$

$V_S=V_{SR}+V_R$

$V_S=(V_R-V_{SR}Sin{30}^{\circ }).\left(0.05\right)$

Displacement along $x-axis$=

$d=(V_R-V_{S}RSin{37}^{\circ }).\left(0.05\right)$

$=[4-5(\frac{3}{5}\left)\right].\left(0.05\right)$

=0.05km

=50m

journey time (only while swimming)

= $\frac{200\times {10}^{-3}}{V_{VR}cos{37}^{\circ }}$

= 0.05hr

total time = $\frac{50\times {10}^{-3}km}{3km/hr}+0.05hr$

= $\frac{0.05}{3}+0.05=\frac{0.20}{3}=0.0666hr=4mins$