An aricraft is flying at 450 ft/s with an elevation of 155 ft from the ground, on a straight-line path that will take it directly over an anti-aircraft gun. How fast, in radians per second, will the gun have to turn to accurately track the aircraft when the plane is:
An picture is given by the question.
1000 feet away = ______ 100 feet away = ______
Please help! I have no idea where to start. Thank you!
$\endgroup$ 51 Answer
$\begingroup$Establish the equation below
$$\tan\theta= \frac hx $$
Take the time derivatives of both sides
$$\sec^2\theta \frac{d\theta}{dt}= -\frac{h}{x^2} \frac{dx}{dt} $$
Then, use $\sec \theta= \sqrt{h^2+x^2}/x$ to simplify the above expression
$$\frac{d\theta}{dt}= -\frac{h}{x^2+h^2} \frac{dx}{dt} $$
Now, plug in the givens, i.e. $dx/dt= -450$, $h=155$ and $x=1000$ to obtain the rate of change for the angle $d\theta/dt$.
At 1000 ft away, the rate of angle change is
$$\frac{d\theta}{dt}= -\frac{155}{1000^2+155^2} (-450)=0.068 \>\text{rad/s} $$
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