Question

Airplanes A and B are flying with constant velocity in same vertical plane at angle 30 degree and 60degree wit respect to horizontal respectively .The speed ofA is 100root2 m/s .At time tis equal to 0s,an observer in A finds B at a distance of 500m .The observer sees B moving with constant velocity perpendicular to the line of motion of A. If at t is equal to t0 ,A just escapes being hit by B,t0 in seconds is______

Answer 1

Answer:

5s

Step-by-step explanation:

Sorces : Correction : speed of A is 100 root(3)m/s, with Figure Attached

1) We have,

Let v=speed of plane B.

Velocity of plane, A and B.

$\vec{v_A}=100\sqrt{3}(cos(30^{\circ})i+sin(30^{\circ})j)\\ \\\vec{v_B}=v(cos(60^{\circ})i+sin(60^{\circ})j)=v(\frac{1}{2}i +\frac{\sqrt{3} }{2}j )$

2)Since, we have given

the oberver in A sees  B moving with constant velocity perpendicular to the line of motion of A.

=> Velocity of B with respect to A is perpendicular to velocity of A

That is,

$\vec{v_{BA}}.\vec{v_A}=0$

Now,

$\vec{v_{BA}}=\vec{v_B}-\vec{v_A}=v(\frac{1}{2}i+\frac{\sqrt{3} }{2}j)-100\sqrt{3}(cos(30^{\circ})i+sin(30^{\circ})j)\\ \\=(\frac{v}{2}-150 )i+(\frac{v\sqrt{3} }{2}-50\sqrt{3})j$

3) We have,

$\vec{v_{BA}}.\vec{v_A}=0\\ \\=>[(\frac{v}{2}-150 )i+(\frac{v\sqrt{3} }{2}-50\sqrt{3})j].[150i+50\sqrt{3}j]=0\\ \\=>3v-4*150=0\\ \\=>v=200m/s$

4) Again,

$\vec{v_{BA}}=(\frac{200}{2}-150 )i+(\frac{200\sqrt{3}}{2}-50\sqrt{3})j\\ \\=(-50i+50\sqrt{3}j)\\ \\=>v_{BA}=\sqrt{50^2+(50\sqrt{3})^2}=100m/s$

5)At t=t_0, A just escapes from being hit.

Relative Distance between them =500m

Relative Speed =100m/s

=>Time =Relative Distance /Relative Speed =500/100=5s