Question

A motorboat is racing towards north at $25 \,km\, h^{-1}$ and the water current in that region is $10\, km\, h^{-1}$ in the direction of $60^°$ east of south. The resultant velocity of the boat is

• A. $11\, km\, h^{-1}$
• B. $22\, km\, h^{-1}$
• C. $33\, km\, h^{-1}$
• D. $44\, km\, h^{-1}$

Answer 1

Answer: (B)

The velocity of the motor boat and the velocity of the water current are represented by vectors $\vec{v}_{b}$ and $\vec{v}_{c}$ as shown in the figure.

Here, $\theta = 180^° - 60^° = 120^°$
$v_b = 25\, km\, h^{-1}$,
$v_c = 10\, km\, h^{-1}$
$\therefore $ According to parallelogram law of vector addition, the magnitude of the resultant velocity of the boat is
$v_{R}=\sqrt{v^{2}_{b}+v^{2}_{c}+2v_{b}v_{c}\,cos\,120^{°}}$
$=\sqrt{\left(25\right)^{2}+\left(10\right)^{2}+2\left(25\right)\left(10\right)\left(\frac{-1}{2}\right)}$
$=\sqrt{625+100-250}$
$ \approx 22\,km\,h^{-1}$