Airplanes A and B are flying with constant velocity in the same vertical plane at angles 30- and 60- with respect to the horizontal respectively as shown in figure- The speed of A is 100-3 ms-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 - t0- A just escapes being hit by B- t0 in seconds is

The relative velocity of B with respect to A is perpendicular to #160;line #160;of motion of A. ∴ V_B #160;cos 30^∘ = V_A ⇒ V_B = 200 m/s And ti ...

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Standard XII

Physics

1st Equation of Motion

Airplanes A a...

Question

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 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, $t_{0}$ in seconds is

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Solution

The relative velocity of B with respect to A is perpendicular to line of motion of A.
$∴ V_{B} c o s 30^{\circ} = V_{A}$
$\Rightarrow V_{B} = 200 m / s$
And time, $t_{0} = \frac{R e l a t i v e d i s t a n c e}{R e l a t i v e v e l o c i t y}$ $= \frac{500}{V_{B} s i n 30^{\circ}} = 5 s e c$

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The relative velocity of B with respect to A is perpendicular to line of motion of A.∴VBcos30∘=VA⇒VB=200m/sAnd time, t0=Relative distanceRelative velocity=500VBsin30∘=5 sec

The relative velocity of B with respect to A is perpendicular to line of motion of A.
$∴ V_{B} c o s 30^{\circ} = V_{A}$
$\Rightarrow V_{B} = 200 m / s$
And time, $t_{0} = \frac{R e l a t i v e d i s t a n c e}{R e l a t i v e v e l o c i t y}$ $= \frac{500}{V_{B} s i n 30^{\circ}} = 5 s e c$