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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)

14 Points
one month ago

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

$t= \frac{6.828l}{v}$