Name: Inertial Frames of Reference
Uploaded: 2019-07-19
Duration: 7 min 10 s
Description: Imagine if you are on a bus or a train, can you say you are in an inertial frame of reference? If yes, how can you predict it? Let us explore more in this vi...

Question

Two boats

A

and

B

moved away from a buoy anchored in the middle of a river.

A

moved along the river and

B

at right angle to it. Having moved off equal distances from the buoy, the boats returned. Find the ratio of the times of motion of the boats, if the velocity of each boat with respect to still water is

η

times greater than the velocity of water current.

Accepted Answer

Let speed of river is $u$ and distance travelled by both the boats be $d$ .

That is, ${\vec{v}}_{r} = u \hat{i}$

Also given, the velocity of each boat with respect to still water is $η$ times the velocity of river.

That is, $v_{A r} = v_{B r} = η u$ .

Time taken by boat $A$ to move along the river,

$t_{A} = \frac{d}{η u + u} + \frac{d}{η u - u}$

$\Rightarrow t_{A} = \frac{2 d η}{(η^{2} - 1) u} \dots (i)$

From the vector diagram of $v_{B r}$ , ${\vec{v}}_{B r} = - v_{B r x} \vec{i} + v_{B r y} \vec{j}$ .

Since the boat $B$ moves right angle to the river flow, velocity of boat with respect to ground,

${\vec{v}}_{B} = 0 \hat{i} + v_{B} \vec{j} = v_{B} \vec{j}$

As ${\vec{v}}_{B r} = {\vec{v}}_{B} - {\vec{v}}_{r}$ ,

$\Rightarrow {\vec{v}}_{B} = {\vec{v}}_{B r} + {\vec{v}}_{r}$

$\Rightarrow {\vec{v}}_{B} = (- v_{B r x} \vec{i} + v_{B r y} \vec{j}) + u \hat{i}$

$\Rightarrow 0 \hat{i} + {\vec{v}}_{B} \hat{j} = (u - v_{B r x}) \vec{i} + v_{B r y} \vec{j}$

Thus, $u - v_{B r x} = 0$

$\Rightarrow v_{B r x} = u$ and $v_{B r y} = v_{B}$ .

As, $v_{B r}^{2} = v_{B r x}^{2} + v_{B r y}^{2}$ ,

$\Rightarrow v_{B}^{2} = v_{B r y}^{2} = v_{B r}^{2} - v_{B r x}^{2}$

$\Rightarrow v_{B}^{2} = {(η u)}^{2} - u^{2}$

$\Rightarrow v_{B} = \sqrt{{(η u)}^{2} - u^{2}}$

Time taken by boat $B$ to move right angle to the river,

$t_{B} = \frac{2 d}{v_{B}} = \frac{2 d}{\sqrt{{(η u)}^{2} - u^{2}}}$

$\Rightarrow t_{B} = \frac{2 d}{u \sqrt{(η^{2} - 1)}} \dots (ii)$

From equation $(i) and (ii)$ ,

$\Rightarrow \frac{t_{A}}{t_{B}} = \frac{η}{\sqrt{η^{2} - 1}}$

Important Questions on Kinematics I

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