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° with the positive direction of X-axis as shown in the figure. The relative velocity of B with respect to that of A is

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° 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√3 m/s along Y-axis (perpendicular to X-axis) (C) 10√5 along the bisection of the velocities of A and B (D) 30 m/s along negative X-axis

Tardigrade Admin · Accepted Answer

The component of velocity of B along x-direction v B x=20 cos 60°=20 × (1/2)=10 m / s v A=10 hat i v B=10 hat i +10 √3 hat j v B A= v B- v A=10 hat i +10 √3 hat j -10 hat i =10 √3 hat j

$10 m / s$ along X-axis

$103$ m/s along Y-axis (perpendicular to X-axis)

$105$ along the bisection of the velocities of A and B

$30 m / s$ along negative X-axis

The component of velocity of $B$ along $x$ -direction
$v_{B x} = 20 cos 6 0^{\circ} = 20 \times \frac{1}{2} = 10 m / s$
$v_{A} = 10 \hat{i}$
$v_{B} = 10 \hat{i} + 103 \hat{j}$
$v_{B A} = v_{B} - v_{A} = 10 \hat{i} + 103 \hat{j} - 10 \hat{i}$
$= 103 \hat{j}$

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

10m/s along X-axis

103​ m/s along Y-axis (perpendicular to X-axis)

105​ along the bisection of the velocities of A and B

30m/s along negative X-axis

The component of velocity of B along x-direction vBx​=20cos60∘=20×21​=10m/s vA​=10i^ vB​=10i^+103​j^​ vBA​=vB​−vA​=10i^+103​j^​−10i^ =103​j^​

$10 m / s$ along X-axis

$103$ m/s along Y-axis (perpendicular to X-axis)

$105$ along the bisection of the velocities of A and B

$30 m / s$ along negative X-axis

The component of velocity of $B$ along $x$ -direction
$v_{B x} = 20 cos 6 0^{\circ} = 20 \times \frac{1}{2} = 10 m / s$
$v_{A} = 10 \hat{i}$
$v_{B} = 10 \hat{i} + 103 \hat{j}$
$v_{B A} = v_{B} - v_{A} = 10 \hat{i} + 103 \hat{j} - 10 \hat{i}$
$= 103 \hat{j}$