#### A motorboat is racing towards north at 25 km/h and the water current in that region is 10 km/h in the direction of 60° east of south. Find the resultant velocity of the boat.

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vector addition using the laws of sines and cosines
law of sines
angle of the resultant vector
motion in a plane
ncert physics
MOTION IN A PLANE

An0nym0us

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Hasini

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**Method of parallelogram:**

If the adjacent sides of a parallelogram drawn from a point can represent two simultaneous vectors, both in magnitude and direction, the resulting vector is represented by the diagonal of a parallelogram which passes across that point, in both magnitude and direction.

Step 1: Draw a figure from the given information

Velocity of the motorboat A = v_b = 25 km/s

The rate of movement in the water B = v_c = 10 km/h

Resultant velocity of the motorboat = R

Angle between the v_b \text{ and } v_c, \theta = 180 - 60 = 120 \degree

Step 1: Obtain the magnitude of the resultant velocity

The Law of cosines:

Magnitude of the resultant vector R = \sqrt{A^2 + B^2 + 2 AB \cos \theta}

R = \sqrt{v_b^2 + v_c^2 + 2 v_b v_c \cos \theta}

R = \sqrt{25^2 + 10^2 + 2 *25*10 \cos 120\degree}

R = \sqrt{625 + 100 + 500 * - \frac{1}{2}}

R = 21.8 Km/h

Step 2: Find the direction of the resultant vector

The Law of sines:

Direction of the resultant vector \sin\phi=\frac{v_c}{R}\sin\theta

[math] \phi = \sin^{-1} [\frac{10}{21.8} \sin 120\degree] [/math]

[math] \phi = \sin^{-1} [\frac{10}{21.8} * \frac{\sqrt{3}}{2} [/math]

[math] \phi = \sin^{-1} [0.397] [/math]

\phi = 23.39\degree

Thus, Direction of the resultant vector \phi = 23.39\degree