Question

A swimmer swims in still water at a speed $= 5 km/hr$. He enters a $200 m$ wide river, having river flow speed $= 4 km/hr$ 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 $= 3 km/hr$. The total time in which he reaches from $A$ to $B$ is
(A) $5 min$
(B) $4 min$
(C) $3 min$
(D) None

Answer 1

Hint
We use here the simple formula ${\text{Distance = Speed}} \times {\text{Tme}}$ to find time in horizontal motion of swimmer and vertical motion of swimmer in the river by dividing the components that is $\operatorname{Sin} \theta {\text{ and }}\operatorname{Cos} \theta$ where $\theta$ is the given angle in which swimmer swims into the river.

Complete step by step solution
Given, swimmer swims in still water at a speed of $5km/hr$ angle of swimming in river = ${127^ \circ }$
Therefore speed towards width will be find with the help of components of angle that is
Vertical speed $=$ $5 \times \sin {127^ \circ }$ $= 5 \times (0.8)$ $= 4km/hr$
Time taken to cover distance $= \dfrac{{0.2}}{4}hr = 0.05hr$
Since 1hour has 60 min. Therefore 0.05hour = $0.05 \times 60 = 3\min$ .
Now Horizontal speed = $5 \times \cos {127^ \circ } = 5 \times ( - 0.6) = - 3km/hr$
Net flow of swimmer toward river flow $= 4 - 3 = 1km/hr$.
We know ${\text{Distance = Speed}} \times {\text{Tme}}$
Distance covered in 3 mins = $\dfrac{3}{{60}} \times 1 = \dfrac{1}{{20}}km = 50m$
Now we found that 50 m distance is covered with the speed of 3km/hr
Therefore time take $= \dfrac{{50}}{{1000}} \times \dfrac{1}{3} = \dfrac{1}{{60}}hr = 1\min$.
Now the total time taken to reach point B from A is $3 + 1 = 4\min$ .
Therefore the correct answer is option (B).

Note
Remember the negative sign of speed shows the opposite direction of the swimmer against the flow of the river. Also remember $1\min = \dfrac{1}{{60}}hr$ . Remember the value of $\sin {37^ \circ } = 0.6$ . also $\sin (90 + \theta ) = \cos \theta$.