question_answer

A jet plane flying at a constant velocity v at a height \[h=8\text{ }km\], is being tracked by a radar R located at O directly below the line of flight. If the angle \[\theta \] is decreasing at the rate of \[0.025\text{ }rad/s\], the velocity of the plane when \[\theta =\text{ }60{}^\circ \]is:

A) 1440 km/h       

B)        960 km/h

C) 1920 km/h       

D)        480 km/h

Correct Answer: B

Solution :

[b]

\[x=h\cot \theta \]

\[\therefore \frac{dx}{dt}=h\frac{d\left( \cot \theta  \right)}{dt}\]

\[=h\,\,(-\,\text{cose}{{\text{c}}^{2}}\theta )\,\,\frac{d\theta }{dt}\]

\[\text{or }\,v=h\text{ cose}{{\text{c}}^{2}}\theta \,\,\left( -\frac{d\theta }{dt} \right)\]

\[=(8\times {{10}^{3}})\,\text{cose}{{\text{c}}^{2}}60{}^\circ \times 0.025=2666.67\text{ m/s}\]

\[\text{=}\,\,\text{960 km/h}\]