Find the time taken to cross the train moving with velocity and calculate minimum time required to cross the river for minimum drift.
\(t = \dfrac{d}{v_{Robin/train }}\)
A 5 sec
B 3 sec
C 4 sec
D 2 sec
\(t_{min} = \dfrac{d}{|\vec{v}_{person/river}|}\)
when, \(v_{person/river}\) is the velocity of person with respect to river.
The distance between A and B is known as drift. Here, it is denoted by \('x'\).
Since, \(\vec{v}_{m/_r}=\;\vec{v}_{m/_G}-\;\vec{v}_{r/_G}\)
Thus, \(t_{min} = \dfrac{Displacement \; in \; y-direction }{Velocity \; in \; y-direction}\)
\(t_{min} = \dfrac{d}{v_{m/r}}\)
The velocity in x-direction is \(\vec{v}_{r/_G}\).
Thus, displacement in x-direction = \(\vec{v}_{r/_G}× t_{min}\)
Drift = \(\vec{v}_{r/_G} × t_{min}\)
\(\vec{v}_{m/_G}=\vec{v}_{m/r}+\vec v_{r/_G}\)
\(\because\;\;\vec{v}_{m/r} = v_{m/r} \,sin \,\theta\,(-\hat i) + v_{m/r}\; cos \;\theta\; \hat j\)
\(\therefore\;\;\vec{v}_{m/_G} = v_{r/_G} \hat i - v_{m/r}\; sin\;\theta \;\hat i + v_{m/r}\; cos \; \theta\, \hat j\)
\(\vec{v}_{m/_G} =\left(v_{r/_G }-v_{m/r} sin \; \theta\; \right)\hat i + \left(v_{m/r\; }cos\; \theta\right) \hat j\)
Thus, for no drift in horizontal direction,
\(v_{r/_G} - \,v_{m/r} \,sin \;\theta \; = 0\)
\(sin \;\theta = \dfrac{v_r}{v_{m/r}}\)
\(\theta = sin ^{-1} \left(\dfrac{v_r}{v_m/_r}\right)\)
\(\vec{v}_{m}=\vec{v}_{m/r}+\vec v_{r}\)
\(\because\;\;\vec{v}_{m/r} = -v_{m/r} \,sin\, \theta\,\hat i + v_{m/r}\; cos \;\theta\; \hat j\)
\(\therefore\;\;\vec{v}_{m} = v_{r} \hat i - v_{m/r}\; sin\;\theta \;\hat i + v_{m/r}\; cos \; \theta \hat j\)
\(\vec{v}_{m} =\left(v_{r } -v_{m/r} sin \; \theta\; \right)\hat i + \left(v_{m/r\; }cos\; \theta\right) \hat j\)
\(\vec{v}_m = (v_m)x\;\hat i + (v_m) y\; \hat j\)
\(\Rightarrow\;\; x = (v_r-v_{m/r}\; sin \theta) × \; \dfrac{width \; of \; river }{velocity \; of \; man \; in \; y\; direction }\)
Thus, minimum drift possible is zero.
\((v_r - v_{m/r}\; sin \; \theta) \dfrac{d}{v_{m/r}\; cos \; \theta} = 0\)
\(\Rightarrow\;\;v_r - v_{m/r} \; sin\, \theta = 0 \)
\(\Rightarrow \;\;sin\, \theta = \dfrac{v_r}{v_{m/r}}\)
\(t = \dfrac{d}{v_{m/r}\; cos \,\theta}\;—\;\;(1)\)
\(cos \,\theta\; = \dfrac{\sqrt {v^2_{m/r} - v_r^2}}{v_{m/r}}\)
Putting the value of \( cos \,\theta \) in equation \(\;(1)\)
Time to cross the river with minimum drift
\(t = \dfrac{d}{\sqrt {v^2_{m/r} - v_r^2}}\)
A 20 minutes
B 10 minutes
C 12 minutes
D 5 minutes
\(\vec v_{m} = (v_r-v_{m/r} \; sin\, \theta ) \hat i + v_{m/r} \; cos \,\theta \;\hat j \)
\(x = (v_r-v_{m/r}\; sin\,\theta)\dfrac{d}{v_{m/r}\; cos \,\theta}\)
As, \(|\vec v_r|>|\vec v _{m/r}|\) thus, drift can't be zero. Because, \(sin \; \theta\) always vary from –1 to 1.
When, \(sin \; \theta\) is –1 then \([v_r + v_{m/r}] \Rightarrow\) drift is non-zero .
When, \(sin \; \theta\) is +1 then \([v_r - v_{m/r}] \Rightarrow\) drift is non-zero.
For minimum value of \(x,\; \dfrac{dx}{d\theta} \; = 0\)
\(\therefore\;\;\dfrac{d}{d \theta} \left(\dfrac{d\;v_r}{v_{m/r}\; cos \,\theta}\right) - \dfrac{d}{d\theta} \left[\dfrac{d\;v_{m/r }\;sin\, \theta}{v_{m/r}\; cos \,\theta}\right] = 0\)
\(\dfrac{dv_r}{v_{m/r}}\;\dfrac{d}{d\theta} (sec\; \theta) - d \left(\dfrac{d}{d\theta} tan\; \theta\right) = 0\)
\(d\left[\dfrac{v_r}{v_{m/r}}sec \,\theta \,tan\,\theta - sec^2\theta\right] = 0\)
\((sec\,\theta)(d) \left[\dfrac{v_r}{v_{m/r}}\left(\dfrac{sin\,\theta}{cos\theta}\right) - \dfrac{1}{cos\, \theta}\right] = 0\)
\(\Rightarrow\;\;\dfrac{v_r\;\;sin\,\theta}{v_{m/r}\;\;cos\,\theta} = \dfrac{1}{cos\,\theta}\)
\(\Rightarrow\;\;sin\,\theta = \dfrac{v_{m/r}}{v_r}\)
Thus, the angle required to cross the river with minimum drift when speed of river is greater than speed of man is given by
\(\theta = sin ^{-1} \left(\dfrac{v_{m/r}}{v_r}\right)\)