Flag Mechanics> A man, running, on the road AA ’, with sp...
question mark

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)

Joshika Kolli , 3 Years ago
Grade 12
anser 1 Answers
Satya Prakash Yadav

Last Activity: 3 Years 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}

Provide a better Answer & Earn Cool Goodies

Enter text here...
star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments


Ask a Doubt

Get your questions answered by the expert for free

Enter text here...