Question

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

• A. $22\text{km/hr}$, at an angle ${\tan }^{-1}\left(0.343\right)$ North of East.
• B. $22\text{km/hr}$, at an angle ${\tan }^{-1}\left(0.433\right)$ East of North.
• C. $20\text{km/hr}$, at an angle ${\tan }^{-1}\left(0.433\right)$ East of North.
• D. $20\text{km/hr}$, at an angle ${\tan }^{-1}\left(0.343\right)$ North of East.

Answer 1

The correct option is B $22\text{km/hr}$, at an angle ${\tan }^{-1}\left(0.433\right)$ East of North.

The vector $v_b$ representing the velocity of the motorboat and the vector $v_c$ representing the water current are shown in the figure in directions specified by the problem.


Using the parallelogram method of addition, the resultant $R$ is obtained in the direction shown in the figure.
So, the magnitude of the resultant is given by
$R=\sqrt{v_b^2+v_c^2+2v_b{v}_c\cos \theta }$, where $\theta ={120}^{\circ }$
$\Rightarrow R=\sqrt{{25}^2+{10}^2+2\times 25\times 10\cos \left({120}^{\circ }\right)}$
$R=21.8\approx 22\text{km/hr}$
Now, for direction we have
$\tan \varphi =\frac{v_c\sin \left({120}^{\circ }\right)}{v_b+v_c\cos \left({120}^{\circ }\right)}=\frac{10\sin \left({120}^{\circ }\right)}{25+10\cos \left({120}^{\circ }\right)}$
$\Rightarrow \tan \varphi =0.433\Rightarrow \varphi ={\tan }^{-1}\left(0.433\right)$ East of North.