Question

Let us consider a boat that moves with a velocity

v_{b, w} = 5 km

h^{- 1}

relative to water. At time

t = 0

, the boat passes through a piece of cork floating in the water while moving downstream. If it turns back at time

t = t_{1}

, when and where does the boat meet the cork again? Assume

t_{1} = 30 \min

.

Accepted Answer

Time of travelling of boat from $A$ to $B (t_{1})$ and then $B$ to $h^{- 1}$ $(t^{'}) =$ time of moving the cork from $A$ to $C$ .

Let the speed of water is $u$ . The velocity of boat from $A$ to $B$ $= {\vec{v}}_{b, w} + {\vec{v}}_{w} = (5 + u) km h^{- 1}$

and velocity of boat from $B$ to $C$ $= {\vec{v}}_{b, w} - {\vec{v}}_{w} = (5 - u) km h^{- 1}$ .

Distance moved by boat in time $t_{1}$

$A B = (5 + u) t$

and distance moved by boat in time $t_{1}^{'} = B C = (5 - u) t_{1}^{'}$ .

Distance moved by cork during this time
$A C = u (t_{1} + t_{1}^{'})$

But, here $A B = A C + B C$

$(5 + u) t_{1} = u (t_{1} + t_{1}^{'}) + (5 - u) t_{1}^{'}$

$5 t_{1} = 5 t_{1}^{'} \Rightarrow t_{1}^{'} = t_{1} = 30 \min$
Hence, the cork meets the boat again after $(t_{1} + t_{1}^{'}) = 1 h$ .

B M Sharma Solutions for Chapter: Kinematics I, Exercise 53: ILLUSTRATION 4.50

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics I, Exercise 53: ILLUSTRATION 4.50

Questions from B M Sharma Solutions for Chapter: Kinematics I, Exercise 53: ILLUSTRATION 4.50 with Hints & Solutions

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