Question

a man swim across a river with speed of Vm perpendiculuar to the flow of direction of river .if water flows with the speed of Vw ,then what is direction of man to cross river ...................how we place theta in this question.....how we know we hv to place theta either in x or y diection ......with resultant velocity pls tell me

Answer 1

Answer on Question#39645 – Physics - Mechanics | Kinematics | Dynamics

a man swim across a river with speed of Vm perpendicular to the flow of direction of river. if water flows with the speed of Vw, then what is direction of man to cross river ...how we place theta in this question...how we know we hv to place theta either in x or y direction ...with resultant velocity pls tell me

Solution:

$V_{m}$ - velocity of the man relative to the water;

$V_{w}$ - velocity of the current;

First, we can find the angle $\theta$ between the direction of velocity of the water and velocity of the man ( $\vec{V}_m$ and $\vec{V}_w$ form a right-angled triangle because $\vec{V}_m$ is perpendicular to the flow):

$$\text {t r i a n g l e} X - \text {d i r e c t i o n :} \tan \theta = \frac {V _ {m}}{V _ {w}} \Rightarrow$$ $$\theta = \arctan \left(\frac {V _ {m}}{V _ {w}}\right)$$ $$Y - \text {d i r e c t i o n :} \sigma = 9 0 {}^ {\circ} - \theta = 9 0 {}^ {\circ} - \arctan \left(\frac {V _ {m}}{V _ {w}}\right)$$

Also we can find resultant velocity using Pythagoras' theorem:

$$\begin{array}{l} \vec {V} _ {\text {r e s u l t}} = \vec {V} _ {m} + \vec {V} _ {w} \\ V _ {\text {r e s u l t}} ^ {2} = V _ {m} ^ {2} + V _ {w} ^ {2} \Rightarrow \\ V _ {r e s u l t} = \sqrt {V _ {m} ^ {2} + V _ {w} ^ {2}} \\ \end{array}$$

Answer: angle with x-direction: $\theta = \arctan \left(\frac{V_m}{V_w}\right)$ ;

y-direction: $\sigma = 90{}^{\circ} - \arctan \left(\frac{V_m}{V_w}\right)$ ;

resultant velocity: $V_{\text{result}} = \sqrt{V_m^2 + V_w^2}$ .