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A man, running, on the road AA ’, with speed v , wants to reach his friend B trapped in the mudfield as early as possible. The speed of man in mudfield is one third of his running speed along AA ’. The minimum time taken (from A to B )

A man, running, on the road AA’, with speed v, wants to reach his friend B trapped in the mudfield as early as possible. The speed of man in mudfield is one third of his running speed along AA’. The minimum time taken (from A to B)
 

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Grade:12

1 Answers

Satya Prakash Yadav
14 Points
one month ago
Answer: (4+2√2)l/v = 6.828(l/v)
 
let us assume man A will leave the road x distance before point D.
length covered by man A on the road will be (4l-x)
 
where r= √(x2 + l2 )
 
total time taken by man A will be 
 
 
t= \frac{4l-x}{v}+\frac{r}{v/3}
 
on differentiatiing w.r.t. x we get,
 
 
\frac{dt}{dx}= \frac{-1}{v}+\frac{3x}{v\sqrt{x^2 + l^2}}
 
for minimum time  \frac{dt}{dx}=0
 
 
and \frac{d^2t}{dx^2}>0
 
 
we get the value of 
 
 
                                 x= \frac{l}{2\sqrt{2}}
 
 
 
 
 
therefore minimum time will be 
 
t= \frac{4l}{v}- \frac{l}{2\sqrt{2}v}+\frac{9l}{2\sqrt{2}v}
 
on calulating we will get 
 
t= \frac{4+2\sqrt{2}}{v}l
that is minimum time will be
 
 
                    t= \frac{6.828l}{v}

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