A boat going down stream crosses a float at a point $A$ . $^{'} t_{1}^{'}$ minutes later the boat reverses its direction and in the next $^{'} t_{2}^{'}$ minutes it crosses the float at a distance $L$ from the point $A$ . The velocity of the river is:

\frac{2 L}{t_{1}}

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\frac{L}{t_{1} + 2 t_{2}}

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\frac{2 L}{2 t_{1} + t_{2}}

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\frac{L}{2 t_{1}}

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Solution

The correct option is D $\frac{L}{2 t_{1}}$
$(D)$
Let $v_{b} :$ velocity of boat
$v_{r} :$ velocity of river
In ground frame
for float : $L = v_{r} (t_{1} + t_{2}) \dots (1)$
for boat : $L = (v_{b} + v_{r}) t_{1} - (v_{b} - v_{r}) t_{2}$
$L = v_{r} (t_{1} + t_{2}) + v_{b} (t_{1} - t_{2}) \dots (2)$
solving $(1) & (2)$
$t_{1} = t_{2}$
From (1) $v_{r} = \frac{L}{t_{1} + t_{2}}$

$v_{r} = \frac{L}{2 t_{2}} = \frac{L}{2 t_{1}}$

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A boat going down stream crosses a float at a point A. ′t′1 minutes later the boat reverses its direction and in the next ′t′2 minutes it crosses the float at a distance L from the point A. The velocity of the river is:

The correct option is D L2t1(D)Let vb: velocity of boat vr: velocity of river In ground frame for float : L=vr(t1+t2)…(1)for boat : L=(vb+vr)t1−(vb−vr)t2 L=vr(t1+t2)+vb(t1−t2) …(2)solving(1) & (2)t1=t2From (1) vr=Lt1+t2vr=L2t2=L2t1

A boat going down stream crosses a float at a point $A$ . $^{'} t_{1}^{'}$ minutes later the boat reverses its direction and in the next $^{'} t_{2}^{'}$ minutes it crosses the float at a distance $L$ from the point $A$ . The velocity of the river is: