Question

A person is rowing across the river with a velocity of 4.5 km/hr northward. The river is flowing eastward at 3.5 km/hr (Figure 4). What is the magnitude of her velocity ( v ) with respect to the shore?

Answer 1

Answer: The magnitude of her velocity ( v ) with respect to the shore is 5.70 km/h.

Explanation:

Magnitude of  Velocity of the boat = Q

Magnitude of Velocity of the river flowing = P

R = Resultant velocity due to velocity of boat and velocity of river.

Applying Law of triangles of  vector addition :

$R=\sqrt{P^2+Q^2+2PQCos\theta}$

From the figure attached:

P = 3.5 k/h, Q = 4.5 km/h

$\theta= 90^o$

$R=\sqrt{P^2+Q^2+2PQCos\theta}=\sqrt{(3.5)^2+(4.5)^2+3.5\times 4.5\times cos90^o}=5.70$

$(Cos90^o=0),(sin 90^o=1)$

$\alpha =tan^{-1}\frac{Qsin\theta}{P+Qcos\theta}=tan^{-1}\frac{4.5 sin 90^o}{3.5+4.5 cos90^o}=tan^{-1}\frac{4.5}{3.5}=52.12^o$

The magnitude of her velocity ( v ) with respect to the shore is 5.70 km/h.