Question

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

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

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, $\theta =180^{°}-60^{°}=120^{°}$
$v_b=25km{h}^{-1}$,
$v_c=10km{h}^{-1}$
$\therefore$ According to parallelogram law of vector addition, the magnitude of the resultant velocity of the boat is
$v_R=\sqrt{v_b^2+v_c^2+2v_b{v}_{c}cos120^{°}}$
$=\sqrt{{\left(25\right)}^2+{\left(10\right)}^2+2\left(25\right)\left(10\right)\left(\frac{-1}{2}\right)}$
$=\sqrt{625+100-250}$
$\approx 22km{h}^{-1}$