Question Of the Day - 21-06-2011 Crossing the river.

1

kunl ·2011-06-20 23:39:35

well clearly in second case its follwoing a parabolic trajectory as far as i can see[1]

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aditya ravichandran ·2011-06-22 08:11:31

an ms-paint version :P

\texttt{The component of man's velocity along radius is responsible for } \\ \texttt{decrease in radius} \\ \\ -\frac{dr}{dt}=v_{m}\sin {\left( \frac{90-\theta}{2}\right)} \cdots\boxed{1}\\ \\ \texttt{The component of man's velocity along tangent is responsible for }\\ \texttt{change in }\theta \\ \\ r\frac{d\theta}{dt}=v_{m}\cos{\left( \frac{90-\theta}{2}\right)} \cdots\boxed{2}\\ \\ \texttt{Dividing both equations} \\ -\frac{dr}{r}=\tan{\left( \frac{90-\theta}{2}\right)}d\theta \cdots\boxed{3}\\ \texttt{Integrating both sides with proper limits} \\ -\int_{d}^{r}\frac{dr}{r}=\int_{0}^{\theta}\tan{\left( \frac{90-\theta}{2}\right)}d\theta \\ \texttt{We finally get} \\ r(1+\sin\theta)=d \\ \texttt{At }\theta=\frac{\pi}{2} \\ \boxed{r=\frac{d}{2}}

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1

aditya ravichandran ·2011-06-22 03:22:56

a purely mathematical proof :

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49

Subhomoy Bakshi ·2011-06-21 13:34:13

Absolute Anand absolute! :D

Waise I knew it! :P

Anyway, officially welcome back! :D

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49

Subhomoy Bakshi ·2011-06-21 13:19:42

1)

v = rÏ‰

2vcosÎ¸ = 2d dÎ¸dt
or,\;\frac{v}{d}\int_{0}^{t}{dt}=\int_{0}^{\frac{\pi }{3}}{sec\theta d\theta }

Gives answer: t=vd log(2+âˆš3)

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1

kunl ·2011-06-21 03:34:56

tht is the picture at t=0 which u have drawn aditya...
now draw the picture at say t=5 second based on the condition that man's velocity is directed towards O...hope u got what why the velocity is not same throughout.

But yeah at every point NET velocity would bisect the angle b/w velocity of man and velocity of river by simple rule of vector algrebraic summation of two vectors[1]

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262

Aditya Bhutra ·2011-06-21 03:23:59

in the second case, why is the velocity changing ??
velocity is same throughout so isnt the path supposed to be str. line ?

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1

kunl ·2011-06-21 03:00:41

btw here's something to admit at TIIT...this is Anand for those who remember me.
And for those who asked my JEE results few days(days?or a month [3]) back but i did not reply here's the reply(actually i was really feeling [2]ashamed[2] of myself for not clearing it even in the second attempt...so did not want to reply[3]):
I failed[2]..tht too really badly not clearing the Chemistry cut-off!

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1

kunl ·2011-06-21 02:57:26

Now can u say at what distance from O would the person end up on the other bank of the river?

since in the post above we proved that it is a parabola so the man would end up at a point on bank where his distane from O and our constructed line is equal(WHY? since e=1 for a parabola)

so,man would end up at a point at a distance of d/2 from O on the other bank of the river when he crosses it[1]

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1

kunl ·2011-06-21 02:54:53

well trying out a GEOMETRIC CONSTRUTION[3] to make things clear.

construct "a line" at a distance d from the man along downstream.

Wondering why?read on,things will justify themselves soon.

At t=0. Man is at a distance d from O and from our line too
At any time t,Man's distance from O and our line would be same since the velocity of man directed towards O=velcoity of man downstream.

so,by now most of us would have got it that the line we constructed turned out to be the so-awaited directrix of parabola(since eccentricity=1) and O is the focus.

Hence we have proved that the trajectory is a parabola[1]