First slide

Relative motion in 2 dimension

Question

A swimmer crosses the river along the line making an angle of 45^o with the direction of flow. Velocity of the river water is 5 m/s. Swimmer takes 12 seconds to cross
the river of width 60 m. The velocity of the swimmer with respect to water will be

Moderate

A

10 m/s
B

5 m/s
C

$5 \sqrt{5} m / s$
D

$5 \sqrt{2} m / s$

Solution

Time taken by man to cross the river
$\begin{array}{l} = \frac{width of river}{v_{y}} \\ 12 = \frac{60}{v_{y}} \Rightarrow v_{y} = 5 m / s \end{array}$
Let the x component of velocity of man w.r.t. river is v_x Since velocity of man w.r.t. ground makes an angle of 45^o with river flow x component of velocity of man w.r.t. ground
= y component of velocity of man w.r.t. ground
$\begin{array}{l} {\vec{v}}_{r} + {\vec{v}}_{x} = {\vec{v}}_{y} \\ v_{x} = 0 \end{array}$
So velocity of man w.r.t. water = v_y= 5 m/s

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