Question

A body moving with 25 ms${}^{-1}$ due east takes a right turn through 60${}^0$ and continues with the same speed. The change in velocity is

• A. $\sqrt{3}\times 25m{s}^{-1}{60}^0$ south of west
• B. 25ms${}^{-1}$ 60${}^0$ south of west
• C. 25ms${}^{-1}$ 60${}^0$ north of east
• D. $\sqrt{2}\times 25m{s}^{-1}$ south of west

Answer 1

Answer: (B)

25ms${}^{-1}$ 60${}^0$ south of west

$\overrightarrow{v_{initial}}=25\hat{i}$
$\overrightarrow{v_{final}}=25(\frac{1}{2}\hat{i}-\frac{\sqrt{3}}{2}\hat{j})$
$\therefore △\overrightarrow{v}=25(-\frac{1}{2}\hat{i}-\frac{\sqrt{3}}{2}\hat{j})$
$\theta ={\tan }^{-1}\frac{\frac{-\sqrt{3}}{2}}{\frac{-1}{2}}$
$\Rightarrow \theta ={60}^0$ and it is south of west.