9.6 Solved examples

  Motion in One Dimension

Question: A river is flowing from west to east at a speed of $5\ m/min$. A man on the south bank of the river capable of swimming at $10\ m/min$ in still water. He wants to swim across the river in shortest time. Find the shortest time and the direction in which he should swim.

Solution: If the man wants to swim across the river in shortest time, then he should swim in the direction perpendicular to the flow of the river i.e. in the north direction.

Question: A river is $500\ m$ wide is flowing at a rate of $3\ m/s$. A boat is sailing at a velocity of $10\ m/s$ with respect to the water, in a direction perpendicular to the river.

(a). Find the time taken by the boat to reach the opposite bank.
(b). How far from the point directly opposite to the starting point does the boat reach the opposite bank?

Solution: Writing all the information below,

${\overrightarrow v _{BR}} = 10\widehat j$
${\overrightarrow v _R} = 3\widehat i$
$d=500\ m$

(a). Since the boat is moving perpendicular to the flow of the river (i.e. $AB$) means the boat is using all of its power to cross the river.

Therefore, the time taken by the boat to cross the river is,
$$t = \frac{{{\text{Width of the river}}}}{{{\text{Velocity perpendicular to the river}}}}$$$$t = \frac{d}{{{v_{BR}}}} = \frac{{500}}{{10}}$$$$t = 50s$$

(b). Drift of the river along the river flow (i.e. $BC$) is,
$${d_r} = {v_R} \times t$$$${d_r} = 3 \times 50$$$${d_r} = 150\,m$$

Question: A river flows at the rate of $3\ km/hr$ and a person can row a boat at a speed of $5\ km/hr$ in still water. If the difference between the times taken to cross the river by the shortest path and the quickest path be $6$ minutes, Find the width of the river.

Solution: Writing all the information as below,

${\overrightarrow v _R} = 3\widehat i$
${v_{BR}} = 5$

For shortest path

Shortest path means directly on the opposite point on the shore.

For no drift we can write,
$${v_R} = {v_{BR}}\sin \theta $$$$\sin \theta = \frac{{{v_R}}}{{{v_{BR}}}} = \frac{3}{5}$$or$$\cos \theta = \frac{4}{5}$$
Similarly, $${v_B} = {v_{BR}}\cos \theta $$Therefore,$${v_B} = 5 \times \frac{4}{5}$$$${v_B} = 4\,km/hr$$ So, the time to cross the river is, $${t_1} = \frac{d}{{{v_B}}}$$$${t_1} = \frac{d}{4}hr\quad ...(i)$$

For quickest path (shortest time).

Time to cross the river is given by,
$${t_2} = \frac{d}{5}hr\quad ...(ii)$$

From question we know, $${t_1} - {t_2} = 6\,\min $$or$${t_1} - {t_2} = \frac{6}{{60}}hr \quad ...(iii)$$
From equation $(i)$, $(ii)$ and $(iii)$ we get,
$$\frac{d}{4} - \frac{d}{5} = \frac{6}{{60}}$$$$\frac{d}{{20}} = \frac{1}{{10}}$$$$d = 2\,km$$

Question: A river is flowing with a speed of $1\ km/hr$. A swimmer wants to go to point $C$ starting from $A$. He swims with a speed of $5\ km/hr$ at an angle $\theta$ w.r.t. the river. Find the value of $\theta$ if $AB=BC$.
Take $\left( {\sin 8^\circ = \frac{1}{{5\sqrt 2 }}} \right)$.

Solution:

The problem can be better visualized as,

Writing all the information below as,

${\overrightarrow v _R} = \widehat i$
$\left| {{{\overrightarrow v }_{MR}}} \right| = 5$

${v_{MR}}$ in the vector form can be written as,

${\overrightarrow v _{MR}} = 5\left( {\cos \theta \widehat i + \sin \theta \widehat j} \right)$

From the concept of relative motion,
$${\overrightarrow v _{MR}} = {\overrightarrow v _M} - {\overrightarrow v _R}$$$${\overrightarrow v _M} = {\overrightarrow v _{MR}} + {\overrightarrow v _R}$$$${\overrightarrow v _M} = 5\left( {\cos \theta \widehat i + \sin \theta \widehat j} \right) + \widehat i$$$${\overrightarrow v _M} = \left( {5\cos \theta + 1} \right)\widehat i + 5\sin \theta \widehat j$$

Since, $$BC=AB$$So, $${v_{{M_x}}}t = {v_{{M_y}}}t$$$$\left( {5\cos \theta + 1} \right)t = 5\sin \theta t$$$$5\cos \theta + 1 = 5\sin \theta $$$$5\left( {\sin \theta - \cos \theta } \right) = 1$$$$\sin \theta - \cos \theta = \frac{1}{5}$$$$\sqrt 2 \left( {\sin \theta \times \frac{1}{{\sqrt 2 }} - \cos \theta \times \frac{1}{{\sqrt 2 }}} \right) = \frac{1}{5}$$$$\sqrt 2 \sin \left( {\theta - 45^\circ } \right) = \frac{1}{5}$$$$\sin \left( {\theta - 45^\circ } \right) = \frac{1}{{5\sqrt 2 }}$$$$\sin \left( {\theta - 45^\circ } \right) = \sin 8^\circ $$$$\theta - 45^\circ = 8^\circ $$$$\theta = 53^\circ $$