Question

A particle moves towards east with velocity $5 \,m/s$. After $10$ seconds its direction changes towards north with same velocity. The average acceleration of the particle is

• A. Zero
• B. $\frac{1}{\sqrt{2}} m/s^2 \,N-W$
• C. $\frac{1}{\sqrt{2}} m/s^2 \,N-E$
• D. $\frac{1}{\sqrt{2}} m/s^2 \,S-W$

Answer 1

Answer: (B)

$\Delta v = 2v \,\sin (\frac{\theta}{2}P) = 2\times 5 \times \sin\,45^{\circ} $
$= \frac{10}{\sqrt{2}}$
$\therefore a = \frac{\Delta v}{\Delta t} $
$= \frac{10/\sqrt{2}}{10} = \frac{1}{\sqrt{2}} m/s^2$