Airplanes A and B are flying with constant velocity in the same vertical plane at angles and with respect to the horizontal respectively as shown in figure. The speed of A is . At time , an observer in A finds B at a distance of . The observer sees B mo

Question

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 $100 \sqrt{3} m/s$ . At time $t = 0 s$ , an observer in A finds B at a distance of $500 m$ . The 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 [2014]

Smart Achievers · Accepted Answer

(5)

$60 °$

$So, v_{A} = v_{B} \cos 30 ° ...(i)$

$Now, {\vec{V}}_{B A} = {\vec{V}}_{B} - {\vec{V}}_{A}$

$= v_{B} \cos 30 ° \hat{i} + v_{B} \sin 30 ° \hat{j} - v_{A} \hat{i}$

$= v_{B} \sin 30 ° \hat{j} [From (i)]$

$= \frac{v_{A}}{\cos 30 °} \sin 30 ° \hat{j} [From (i)]$

$= v_{A} \tan 30 ° \hat{j} = \frac{v_{A}}{\sqrt{3}} = 100 m/s$

$So, t_{0} = \frac{500}{| {\vec{V}}_{B A} |} = \frac{500}{100} = 5 sec .$

Solution

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Solution

Ans. (5) As 'A' see the motion of 'B' ⊥r to V→A. So, vA=vBcos30° ...(i) Now, V→BA=V→B-V→A =vBcos30°i^+vBsin30°j^-vAi^ =vBsin30°j^ [From (i)] =vAcos30°sin30°j^ [From (i)] =vAtan30°j^=vA3=100 m/s So, t0=500|V→BA|=500100=5 sec.

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