Question

The velocity of a particle at time $t=0$ is $2m/s$. A constant acceleration of $2m/s^2$ acts on the particle for $2$ seconds at an angle of $60$${}^{\circ }$ with its initial velocity. The magnitude of velocity and displacement of particle at the end of $t=2s$ respectively are:

• A. $2$$\sqrt{7}$ m/s, $4$$\sqrt{3}$ m
• B. $2$$\sqrt{3}$ m/s, $4$$\sqrt{7}$ m
• C. $4$$\sqrt{7}$ m/s, $2$$\sqrt{3}$ m
• D. $4$$\sqrt{3}$ m/s, $2$$\sqrt{7}$ m

Answer 1

The correct option is A $2$$\sqrt{7}$ m/s, $4$$\sqrt{3}$ m

initial velocity $u=2m/s$

acceleration $a$ = $2m/s$ at $60$${}^{\circ }$

component of acceleration in direction of velocity and perpendicular to initiail velocity according to figure

$a_x=acos{60}^0=2\times \frac{1}{2}=1a_y=asin{60}^0=2\times \frac{\sqrt{3}}{2}=\sqrt{3}$

$s_x=ut+\frac{1}{2}{a}_x{t}^2=2\times 2+\frac{1}{2}\times 1{\left(2\right)}^2=6m$

$s_y=ut+\frac{1}{2}{a}_x{t}^2=0+\frac{1}{2}\times \sqrt{3}{\left(2\right)}^2=2\sqrt{3}m$

net displacement = $S=\sqrt{{s_x}^2+{s_y}^2}=\sqrt{36+12}=4\sqrt{3}m$

again for velocity in X any Y direction

$v_x=u+a_{x}t=2+1\times 2=4m/s{v}_y=u+a_{y}t=0+\sqrt{3}\times 2=2\sqrt{3}m/s$

net velocity = $V=\sqrt{{v_x}^2+{v_y}^2}=\sqrt{16+12}=2\sqrt{7}m/s$

thus .

option $A$ is correct