A) \[-\frac{4}{125}rad/s\]
B) \[-\frac{2}{25}rad/s\]
C) \[-\frac{1}{625}rad/s\]
D) None of these
Correct Answer: A
Solution :
[a] Let CD be the position of man at any time t. |
Let BD be x. Then\[EC=x\]. Let \[\angle ACE\] be\[\theta \]. |
Given \[AB=41.6m,CD=1.6m,\] |
and \[\frac{dx}{dt}=2\,m/s\]. |
\[AE=AB-EB=AB-CD=41.6-1.6=40\,m\] |
We have to find \[\frac{d\theta }{dt}\] where\[x=30\,m\]. |
From \[\Delta AEC,\tan \theta =\frac{AE}{EC}=\frac{40}{x}\] |
Differentiating w.r.t. to t, \[{{\sec }^{2}}\theta \frac{d\theta }{dt}=\frac{-40}{{{x}^{2}}}\frac{dx}{dt}\] |
or \[{{\sec }^{2}}\theta \frac{d\theta }{dt}=\frac{-40}{{{x}^{2}}}\times 2\] |
or \[\frac{d\theta }{dt}=\frac{-80}{{{x}^{2}}}{{\cos }^{2}}\theta =-\frac{80}{{{x}^{2}}}\frac{{{x}^{2}}}{{{x}^{2}}+{{40}^{2}}}\] |
\[=-\frac{80}{{{x}^{2}}+{{40}^{2}}}\]. |
When\[x=30m,\frac{d\theta }{dt}=-\frac{80}{{{30}^{2}}+{{40}^{2}}}=-\frac{4}{125}rad/s\]. |
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