Q. The velocity of a body at time $t = 0$ is $10\sqrt{2}\ m/s$ in the north-east direction and it is moving with an acceleration of $2\ m/s^{2}$ directed towards the south. The magnitude and direction of the velocity of the body after $5\ sec$ will be
A) $10\ m/s$, towards east
B) $10\ m/s$, towards north
C) $10\ m/s$, towards south
D) $10\ m/s$, towards north-east
Explanation
(a) $\overrightarrow{v} = \overrightarrow{u} + \overrightarrow{a}t\therefore v = \sqrt{u^{2} + a^{2}t^{2} + 2u\ at\cos\theta}$
$v = \sqrt{200 + 100 + 2 \times 10\sqrt{2} \times 10 \times \cos 135} = 10\ m/s$
$\tan\alpha = \frac{at\sin\theta}{u + at\cos\theta}\frac{10\sin{135{^\circ}}}{10\sqrt{2} + 10\cos{135{^\circ}}} = 1\ \therefore\alpha = 45{^\circ}$
$i.e$., resultant velocity is $10\ m/s$ towards East
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.