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River–boat or river–man problem

Let a boat can travel in still water with a velocity v. If water also moves in the river, then net velocity of boat (wrt ground) will be different from v.
Net velocity of the boat: 32702.png where 32696.png is velocity of boat in water or wrt water and 32690.png is the velocity of water in river.

Special cases

  1. If boat travels downstream:
    Fig. 3
    We have 32678.png and 32672.png, so
    32665.png32657.png = 32651.png
    Hence, the net velocity of the boat will be (v + u) downstream.
  2. If boat travels upstream
    Fig. 4
    We have 32639.png and 32633.png, So
    32627.png32621.png = 32615.png
    (–ve sign indicates backward direction)
    Hence, the net velocity of the boat will be v – u upstream.
  3. If boat travels at some angle θ with downstream:
    Fig. 5
    so 32596.png
    Time taken to cross the river: t = 32584.png
    Drift: x = (u + v cos θ)t  x 32577.png

To cross the river in shortest time

sin θ = 1 ⇒ θ = 90°, tmin­ = 32571.pngx = 32565.png
Fig. 6
x can be calculated from Fig. 6 also: tan α = 32553.png ⇒ x = 32547.png
Net velocity of boat (put θ = 90°), 32541.png,
Magnitude, 32535.png 

To cross the river by shortest path

For this x = 0  u v cos θ = 0
Fig. 7
cos θ = –32523.png  cos(90 + β) = –32517.png  sin β = 32511.png
Net velocity of boat: 32492.png,
Magnitude, 32486.png = 32480.png 32474.png

Rain–man problem

Formula to be applied: 32468.png where 32462.png is velocity of rain wrt man, 32456.png is the velocity of rain (wrt ground) and 32450.png is the velocity of man (w.r.t. ground).
Case I: Rain is falling vertically downwards with velocity vr and a man is running horizontally with velocity vm as shown in Fig. 8. What is the relative velocity of rain wrt man?
Fig. 8
Given: 32438.png,
Now 32432.png 
or 32425.png 
Fig. 9
Magnitude: 32413.png,
Direction: tan α = 32407.png
Case II: If rain is already falling at some angle θ with horizontal (Fig. 10), then with what velocity the man should travel so as to him the rain appears falling vertically downwards?
Fig. 10
Here: 32394.png 
Now 32388.png = (vr sin θ – vm)32382.png
Now for rain to appear falling vertically, the horizontal component of 32376.png should be zero, i.e.,
vr sin θ – vm = 0  sin θ = 32370.png and vr/m = vr cos θ = vr
Fig. 11

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