Airplanes A and B are flying with constant velocity in the same… | MHT CET 2026 (12 May, Shift 1)

Q4MHT CET 2026 · 12 MAY · SHIFT 1Difficult

Relative Velocity in Two Dimensions

Airplanes 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 as shown in the figure. The speed of A is $100\sqrt{3}$ m/s. 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, $t_0$ in second is

$\{\sin(30^\circ) = 0.5, \cos(30^\circ) = 3^{0.5}/2\}$

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Answer: (d)

B is seen moving perpendicular to A's path, so the component of $\vec v_B$ along A's direction equals $v_A$ .
Angle between the two paths is $60^\circ - 30^\circ = 30^\circ$ : $v_B\cos 30^\circ = 100\sqrt{3} \Rightarrow v_B = 200$ m/s.
Relative speed (perpendicular to A's path) $= v_B\sin 30^\circ = 200 \times 0.5 = 100$ m/s.
$t_0 = \dfrac{500}{100} = 5$ s
So the answer is (d) $5$ .

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Relative Velocity in Two Dimensions question from MHT CET 2026 (12 May, Shift 1)

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