Question

Air planes A and B are flying with constant velocity in the same vertical plane at angles 30${}^{\circ }$ and 60${}^{\circ }$ with respect to the horizontal respectively as shown in figure. The speed of A is 100$\sqrt{3}m{s}^{-1}$. At time t = 0 s, an observer in A finds B at a distance of 500 m. This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at t = $t_0$, A just escapes being hit by B, $t_0$ is:

• A. 3 s
• B. 5 s
• C. 2 s
• D. 1 s

Answer 1

Answer: (B)

5 s

The relative velocity of B with respect to A is perpendicular to the line of motion of A.

$\therefore$ $v_{B}cos{30}^{\circ }=v_A{v}_B$= $\frac{v_A}{cos{30}^{\circ }}$

$=\frac{100\sqrt{3}}{\sqrt{3}/2}$ = $200m{s}^{-1}$

$\therefore$ $t_0=\frac{500}{v_{B}sin{30}^{\circ }}$

$=\frac{500}{200X\frac{1}{2}}$

$=5s$