Question

Particle $A$ moves along $X$-axis with a uniform velocity of magnitude $10\, m / s$. Particle $B$ moves with uniform velocity $20\, m / s$ along a direction making an angle of $60^{\circ}$ with the positive direction of $X$-axis as shown in the figure. The relative velocity of $B$ with respect to that of $A$ is

• A. $10\, m/s$ along X-axis
• B. $10\sqrt{3}$ m/s along Y-axis (perpendicular to X-axis)
• C. $10\sqrt{5}$ along the bisection of the velocities of A and B
• D. $30\, m/s$ along negative X-axis

Answer 1

The Correct Option is B

The component of velocity of $B$ along $x$-direction $v _{B x}=20 \cos 60^{\circ}=20 \times \frac{1}{2}=10\, m / s$ $v _{A}=10 \hat{ i }$ $v _{B}=10 \hat{ i }+10 \sqrt{3} \hat{ j }$ $v _{B A}= v _{B}- v _{A}=10 \hat{ i }+10 \sqrt{3} \hat{ j }-10 \hat{ i }$ $=10 \sqrt{3} \hat{ j }$