Kinematics Question 635
Question: A jet plane flying at a constant velocity v at a height $ h=8km $ , it’s being tracked by a radar R located at O directly below the line of flight. If the angle $ \theta $ it’s decreasing at the rate of $ 0.025rad/s $ , the velocity of the plane when $ \theta =60{}^\circ $ it’s:
Options:
A) 1440 km/h
B)960 km/h
C) 1920 km/h
D)480 km/h
Show Answer
Answer:
Correct Answer: B
Solution:
[b]
$ x=h\cot \theta $
$ \therefore \frac{dx}{dt}=h\frac{d( \cot \theta)}{dt} $
$ =h(-\text{cose}{{c^{2}}}\theta )\frac{d\theta }{dt} $
$ \text{or }v=h\text{ cose}{{c^{2}}}\theta ( -\frac{d\theta }{dt} ) $
$ =(8\times 10^{3})\text{cose}{{c^{2}}}60{}^\circ \times 0.025=2666.67\text{ m/s} $
$ \text{=}\text{960 km/h} $
