Question

A jet plane flying at a constant velocity $v$ at a height $h=8$ kilometer 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 $0.025{rads}^{-1}$, the velocity of the plane when $\theta ={60}^o$ is

• A. $1440km{h}^{-1}$
• B. $793.91km{h}^{-1}$
• C. $1920km{h}^{-1}$
• D. $480km{h}^{-1}$

Answer 1

The correct option is B $793.91km{h}^{-1}$

Angle $\varphi$ is increasing at the rate of $0.025rad/sec=\omega$

$\varphi =\omega t$

$t=\frac{\varphi }{\omega }=\frac{\pi }{6}\times \frac{1}{0.025}$

$t=20.94s=0.0058h$

$\tan 30°=\frac{AB}{8}$

$\frac{1}{\sqrt{3}}=\frac{AB}{8}$

$AB=\frac{8}{\sqrt{3}}km$

$Vt=\frac{8}{\sqrt{3}}km$

$V=\frac{8}{\sqrt{3}\times t}$

$V=793.91km/h$