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Motion In Moving Medium

Find the time taken to cross the train moving with velocity and calculate minimum time required to cross the river for minimum drift.

Time taken to Cross the Train Moving with Velocity v

• Consider a train having two gates which is moving with velocity v, as shown in figure.
• Let the distance between gate 1 and 2, is 'd'.
• Suppose, Robin is standing at gate 1.
• He moved from gate 1 and went to gate 2 with some velocity in moving train.
• The time taken by her to cover the distance 'd' is independent of the velocity of train.

• In the reference frame attached to the train, the train will always be at rest and the velocity of Robin with respect to train is $$\text v_{Robin/train }$$ .
• Thus, the velocity of Robin in the reference frame attached with train is equal to the velocity of Robin in stopped train.
• Time taken to cover distance 'd' ,

$$t = \dfrac{d}{v_{Robin/train }}$$

A train is moving with velocity $$v= 3 \,m/s$$ , Ana wants to go across the train from point A to B. If the velocity of Ana when train is not moving is, $$\vec{v}_{Ana/train}= 2 \,m/s$$ and distance between point A and B is d = 8 m, then find the time taken by Ana to reach point B.

A 5 sec

B 3 sec

C 4 sec

D 2 sec

×

Velocity of Ana when train is not moving = Velocity of Ana in reference frame attached to train

Let,  $$\vec{v}_{Ana/train}$$  be the velocity of Ana in reference frame attached to train.

Here, $$|\vec{v}_{Ana/train}|$$ = 2 m/s

Time taken $$=\;\;\dfrac{Distance \;between\;A\;and \; B }{|\vec{v}_{Ana/train}|}$$

$$t= \dfrac{8}{2}= 4 \;sec$$

A train is moving with velocity $$v= 3 \,m/s$$ , Ana wants to go across the train from point A to B. If the velocity of Ana when train is not moving is, $$\vec{v}_{Ana/train}= 2 \,m/s$$ and distance between point A and B is d = 8 m, then find the time taken by Ana to reach point B.

A

5 sec

.

B

3 sec

C

4 sec

D

2 sec

Option C is Correct

Minimum Time Required to Cross the River

• Consider a train moving with velocity v, as shown in figure.

• If a person wants to go from point A to point B in minimum time, then he must cross the train perpendicularly i.e., his velocity with respect to train must be directed perpendicularly to the velocity of train.
• Similarly, if a person wants to cross the river in minimum time, then his velocity with respect to river must be directed perpendicularly to the flow of river, as shown in figure.

• Minimum Time Required to Cross the River

$$t_{min} = \dfrac{d}{|\vec{v}_{person/river}|}$$

when,  $$v_{person/river}$$  is the velocity of person with respect to river.

Robin can swim in still water with velocity  $$\vec{v}_{Robin/current}= 4 \,km/hr$$. If she wants to cross a river of width d = 800 $$m$$ having constant flow of current $$v = 3 \,km/hr$$, then, find the minimum time required to cross the river.

A 12 min

B 15 min

C 20 min

D 30 min

×

Velocity of Robin in still water = Velocity of Robin in the frame attached to river

Let , $$\vec{v}_{Robin/current}$$  is the velocity of Robin with respect to flow of current.

To cross the river in minimum time, Robin has to keep her velocity perpendicular to the flow of current.

Minimum time required,  $$t_{min} = \dfrac{width}{|\vec{v}_{Robin /current}|}$$

$$= \dfrac{0.8}{4} = 0.2 \;hrs\;\;\\= 12 \; min$$

Robin can swim in still water with velocity  $$\vec{v}_{Robin/current}= 4 \,km/hr$$. If she wants to cross a river of width d = 800 $$m$$ having constant flow of current $$v = 3 \,km/hr$$, then, find the minimum time required to cross the river.

A

12 min

.

B

15 min

C

20 min

D

30 min

Option A is Correct

Calculation of Drift

• Consider a swimmer which can swim in still water with some velocity.
• The velocity of swimmer in still water is equal to the velocity of swimmer in reference frame attached to river.
• Let,  $$\vec{v}_{m/r}$$ is the velocity of swimmer(m) with respect to the river.
• Suppose he keeps his velocity perpendicular to the flow of current then he reaches at point B instead of A, as shown in figure.

• Here, $$\vec{v}_{r/_G}$$ is the velocity of river with respect to ground and  $$\vec{v}_{m/_G}$$  is the velocity of man with respect to ground.

• Drift

The distance between A and B is known as drift. Here, it is denoted by  $$'x'$$.

• Calculation of drift

Since, $$\vec{v}_{m/_r}=\;\vec{v}_{m/_G}-\;\vec{v}_{r/_G}$$

Thus, $$t_{min} = \dfrac{Displacement \; in \; y-direction }{Velocity \; in \; y-direction}$$

$$t_{min} = \dfrac{d}{v_{m/r}}$$

The velocity in x-direction is $$\vec{v}_{r/_G}$$.

Thus, displacement in x-direction = $$\vec{v}_{r/_G}× t_{min}$$

Drift = $$\vec{v}_{r/_G} × t_{min}$$

Sara can swim in still water with a velocity of  $$\vec{v}_{Sara/river} = 3\; km / hr$$ . The speed of the current of water is $$\vec{v}_{river/Ground} = 5 \; km/hr$$ . If the width of the river is $$d = 3 \,km$$, then calculate the drift.

A 3 km

B 5 km

C 4 km

D 1 km

×

The velocity of swimmer in still water is equal to the velocity of swimmer in reference frame attached to river.

Let,  $$\vec{v}_{Sara/river}$$  is the velocity of Sara with respect to the flow of current.

To cross the river in minimum time, Sara has to keep her velocity perpendicular to the flow of current.

Minimum time required, $$t_{min} = \dfrac{width}{|\vec{v}_{sara/river}|}$$

$$=\dfrac{3}{3}\\=\; 1 \; hour$$

Drift = $$\vec{v}_{river/Ground } × t_{min}\;\,\,\,\,\,\,\bigg[\because\; Given\,:– \;\vec{v}_{river/Ground }=5\, km/hr\bigg]$$

Drift $$= 5 × 1$$

Drift $$= 5$$ km

Sara can swim in still water with a velocity of  $$\vec{v}_{Sara/river} = 3\; km / hr$$ . The speed of the current of water is $$\vec{v}_{river/Ground} = 5 \; km/hr$$ . If the width of the river is $$d = 3 \,km$$, then calculate the drift.

A

3 km

.

B

5 km

C

4 km

D

1 km

Option B is Correct

Angle Required to Cross the River for Minimum Drift(Vr<Vm/r)

• Consider a river is flowing with velocity  $$v_{r/_G}$$  and a man tries to cross the river such that the drift becomes zero.
• To minimize the drift, the man should swim with an angle $$\theta$$, as shown in figure.

• The angle must be chosen such that there must be no drift in horizontal direction i.e., in positive x-direction.
• For zero drift in horizontal direction, there should be no component of velocity of man in that direction.

$$\vec{v}_{m/_G}=\vec{v}_{m/r}+\vec v_{r/_G}$$

$$\because\;\;\vec{v}_{m/r} = v_{m/r} \,sin \,\theta\,(-\hat i) + v_{m/r}\; cos \;\theta\; \hat j$$

$$\therefore\;\;\vec{v}_{m/_G} = v_{r/_G} \hat i - v_{m/r}\; sin\;\theta \;\hat i + v_{m/r}\; cos \; \theta\, \hat j$$

$$\vec{v}_{m/_G} =\left(v_{r/_G }-v_{m/r} sin \; \theta\; \right)\hat i + \left(v_{m/r\; }cos\; \theta\right) \hat j$$

Thus, for no drift in horizontal direction,

$$v_{r/_G} - \,v_{m/r} \,sin \;\theta \; = 0$$

$$sin \;\theta = \dfrac{v_r}{v_{m/r}}$$

$$\theta = sin ^{-1} \left(\dfrac{v_r}{v_m/_r}\right)$$

A river is flowing at a speed of  $$\vec{v}_{river/Ground} = 3 \; km/hr$$ . Tina can swim in still water with a speed of  $$\vec{v}_{Tina/river} = 5 \; km/h$$ . Calculate the angle $$\theta$$ at which she should swim to cross the river with minimum drift.  $$\left[sin\;37° = \dfrac{3}{5}\right]$$

A 37°

B 47°

C 57°

D 90°

×

The velocity of swimmer in still water is equal to the velocity of swimmer in reference frame attached to river.

As,  $$\vec v_{Tina/river} = 5 \; km/hr$$

and  $$\vec v_{river/Ground} = 3 \; km /hr$$

For minimum drift,

$$\because\;\;\vec{v}_{T/r} = v_{T/r} \,sin \,\theta\,(-\hat i) + v_{T/r}\; cos \;\theta\; \hat j$$

$$\therefore\;\;\vec{v}_{T/_G} = v_{r/_G} \hat i - v_{T/r}\; sin\;\theta \;\hat i + v_{T/r}\; cos \; \theta\, \hat j$$

$$\vec{v}_{T/_G} =\left(v_{r/_G }-v_{T/r} sin \; \theta\; \right)\hat i + \left(v_{T/r\; }cos\; \theta\right) \hat j$$

Thus, for no drift in horizontal direction,

$$v_{r/_G} - \,v_{T/r} \,sin \;\theta \; = 0$$

$$sin \;\theta = \dfrac{v_r}{v_{T/r}}$$

$$\theta = sin ^{-1} \left(\dfrac{v_r}{v_{T/_r}}\right)$$

$$\theta\;=sin^{-1}\left(\dfrac{\vec v_{river/Ground}}{\vec v_{Tina /river}}\right)$$

$$\theta\; = sin ^{-1}\left(\dfrac{3}{5}\right)$$

$$\theta\; = sin ^{-1}( sin \;3 7° )$$

$$\theta = 37 °$$

A river is flowing at a speed of  $$\vec{v}_{river/Ground} = 3 \; km/hr$$ . Tina can swim in still water with a speed of  $$\vec{v}_{Tina/river} = 5 \; km/h$$ . Calculate the angle $$\theta$$ at which she should swim to cross the river with minimum drift.  $$\left[sin\;37° = \dfrac{3}{5}\right]$$

A

37°

.

B

47°

C

57°

D

90°

Option A is Correct

Time Required to Cross the River for Minimum Drift (Vr<Vm/r)

• Consider a river is flowing with velocity $$\vec v_{r/_G}$$ and a man tries to cross the river such that the drift becomes zero.
• To minimize the drift, the man should swim with an angle $$\theta$$, as shown in figure.

• The angle must be chosen such that there must be no drift in horizontal direction i.e., in positive x-direction.
• For zero drift in horizontal direction, there should be no component of velocity of man in that direction.

$$\vec{v}_{m}=\vec{v}_{m/r}+\vec v_{r}$$

$$\because\;\;\vec{v}_{m/r} = -v_{m/r} \,sin\, \theta\,\hat i + v_{m/r}\; cos \;\theta\; \hat j$$

$$\therefore\;\;\vec{v}_{m} = v_{r} \hat i - v_{m/r}\; sin\;\theta \;\hat i + v_{m/r}\; cos \; \theta \hat j$$

$$\vec{v}_{m} =\left(v_{r } -v_{m/r} sin \; \theta\; \right)\hat i + \left(v_{m/r\; }cos\; \theta\right) \hat j$$

$$\vec{v}_m = (v_m)x\;\hat i + (v_m) y\; \hat j$$

• Drift,  $$x = (\vec {v}_m)x × t_{min}\;\;\;\;\;\;\;\;\;\;\left[\because\;(\vec{v}_m)_x = v_r -v_{m/r} \;sin \; \theta\right]$$

$$\Rightarrow\;\; x = (v_r-v_{m/r}\; sin \theta) × \; \dfrac{width \; of \; river }{velocity \; of \; man \; in \; y\; direction }$$

• Since, $$|v_r|\leq|v_{m/r}|$$

Thus, minimum drift possible is zero.

$$(v_r - v_{m/r}\; sin \; \theta) \dfrac{d}{v_{m/r}\; cos \; \theta} = 0$$

$$\Rightarrow\;\;v_r - v_{m/r} \; sin\, \theta = 0$$

$$\Rightarrow \;\;sin\, \theta = \dfrac{v_r}{v_{m/r}}$$

• $$\text {Time required to cross the river} = \dfrac{\text {width of river}}{\text {component of velocity along y-axis}}$$

$$t = \dfrac{d}{v_{m/r}\; cos \,\theta}\;—\;\;(1)$$

$$cos \,\theta\; = \dfrac{\sqrt {v^2_{m/r} - v_r^2}}{v_{m/r}}$$

Putting the value of $$cos \,\theta$$ in equation $$\;(1)$$

Time to cross the river with minimum drift

$$t = \dfrac{d}{\sqrt {v^2_{m/r} - v_r^2}}$$

A river is flowing at a speed  $$\vec{v}_{river/Ground }\; =\; 3 \; km/hr$$. Ron can swim in still water with a speed of $$\vec v_{Ron /river \; }=\; 5 \; km/hr$$. Calculate the time required to cross the river with minimum drift, if the width of river is $$d = 800\, m$$ .

A 20 minutes

B 10 minutes

C 12 minutes

D 5 minutes

×

The velocity of swimmer in still water is equal to the velocity of swimmer in reference frame attached to river.

As,  $$\vec v_{Ron/river} = 5 \; km/hr$$

$$v_{river/Ground }= \;3\; km/hr$$

Width of river, $$d = 800\, m$$

For zero drift in horizontal direction, there should be no component of velocity of Ron in that direction.

$$\vec{v}_{R}=\vec{v}_{R/r}+\vec v_{r}$$

$$\because\;\;\vec{v}_{R/r} = -v_{R/r} \,sin\, \theta\,\hat i + v_{R/r}\; cos \;\theta\; \hat j$$

$$\therefore\;\;\vec{v}_{R} = v_{r} \hat i - v_{R/r}\; sin\;\theta \;\hat i + v_{R/r}\; cos \; \theta \hat j$$

$$\vec{v}_{R} =\left(v_{r } -v_{R/r} sin \; \theta\; \right)\hat i + \left(v_{R/r\; }cos\; \theta\right) \hat j$$

$$\vec{v}_R = (v_R)x\;\hat i + (v_R) y\; \hat j$$

Drift,  $$x = (\vec {v}_R)x × t_{min}\;\;\;\;\;\;\;\;\;\;\left[\because\;(\vec{v}_R)_x = v_r -v_{R/r} \;sin \; \theta\right]$$

$$\Rightarrow\;\; x = (v_r-v_{R/r}\; sin \theta) × \; \dfrac{\text {width of river} }{\text {velocity of man in y direction }}$$

• Since, $$|v_r|\leq|v_{R/r}|$$

Thus, minimum drift possible is zero.

$$(v_r - v_{R/r}\; sin \; \theta) \dfrac{d}{v_{R/r}\; cos \; \theta} = 0$$

$$\Rightarrow\;\;v_r - v_{R/r} \; sin\, \theta = 0$$

$$\Rightarrow \;\;sin\, \theta = \dfrac{v_r}{v_{R/r}}$$

$$\text {Time required to cross the river} = \dfrac{\text {width of river}}{\text {component of velocity along y-axis}}$$

$$t = \dfrac{d}{v_{R/r}\; cos \,\theta}\;—\;\;(1)$$

$$cos \,\theta\; = \dfrac{\sqrt {v^2_{R/r} - v_r^2}}{v_{R/r}}$$

Putting the value of $$cos \,\theta$$ in equation $$\;(1)$$

Time to cross the river with minimum drift

$$t = \dfrac{d}{\sqrt {v^2_{R/r} - v_r^2}}$$

The time required to cross the river with minimum drift is given by

$$t = \dfrac{d}{\sqrt{\vec v\,^{2}_{Ron/ river} - \vec v\;^2_{river/Ground }}}$$

$$t = \dfrac{0.8}{\sqrt{(5)^2-(3)^2}}$$

$$t = \dfrac{0.8}{\sqrt{25-9}}$$

$$t = \dfrac{0.8}{\sqrt{16}}$$

$$t = \dfrac{0.8}{4}$$

$$=0.2\;hr\;\\=12 \;\;\text {minutes}$$

A river is flowing at a speed  $$\vec{v}_{river/Ground }\; =\; 3 \; km/hr$$. Ron can swim in still water with a speed of $$\vec v_{Ron /river \; }=\; 5 \; km/hr$$. Calculate the time required to cross the river with minimum drift, if the width of river is $$d = 800\, m$$ .

A

20 minutes

.

B

10 minutes

C

12 minutes

D

5 minutes

Option C is Correct

Angle Required to Cross the River with Minimum Drift (Vr>Vr/m)

• Consider a river is flowing with velocity $$\vec v_{r/_G}$$ and a man tries to cross the river such that the drift becomes zero.
• To minimize the drift, the man should swim with an angle $$\theta$$, as shown in figure.

• The angle must be chosen such that there must be no drift in horizontal direction i.e., in positive x-direction.
• For zero drift in horizontal direction, there should be no component of velocity of man in that direction.

• $$\vec{v}_{m}=\vec{v}_{m/r}+\vec v_{r}$$

$$\vec v_{m} = (v_r-v_{m/r} \; sin\, \theta ) \hat i + v_{m/r} \; cos \,\theta \;\hat j$$

• Drift, $$x = \vec v_{r/_G}\; × \; t_{min}\;\;\;\;[\because\;v_{r/_G} = v_r - v_{m/r}\; sin\, \theta]$$

$$x = (v_r-v_{m/r}\; sin\,\theta)\dfrac{d}{v_{m/r}\; cos \,\theta}$$

As,  $$|\vec v_r|>|\vec v _{m/r}|$$ thus, drift can't be zero. Because, $$sin \; \theta$$ always vary from –1 to 1.

When, $$sin \; \theta$$ is –1 then $$[v_r + v_{m/r}] \Rightarrow$$ drift is non-zero .

When,  $$sin \; \theta$$  is +1 then $$[v_r - v_{m/r}] \Rightarrow$$ drift is non-zero.

For minimum value of $$x,\; \dfrac{dx}{d\theta} \; = 0$$

$$\therefore\;\;\dfrac{d}{d \theta} \left(\dfrac{d\;v_r}{v_{m/r}\; cos \,\theta}\right) - \dfrac{d}{d\theta} \left[\dfrac{d\;v_{m/r }\;sin\, \theta}{v_{m/r}\; cos \,\theta}\right] = 0$$

$$\dfrac{dv_r}{v_{m/r}}\;\dfrac{d}{d\theta} (sec\; \theta) - d \left(\dfrac{d}{d\theta} tan\; \theta\right) = 0$$

$$d\left[\dfrac{v_r}{v_{m/r}}sec \,\theta \,tan\,\theta - sec^2\theta\right] = 0$$

$$(sec\,\theta)(d) \left[\dfrac{v_r}{v_{m/r}}\left(\dfrac{sin\,\theta}{cos\theta}\right) - \dfrac{1}{cos\, \theta}\right] = 0$$

$$\Rightarrow\;\;\dfrac{v_r\;\;sin\,\theta}{v_{m/r}\;\;cos\,\theta} = \dfrac{1}{cos\,\theta}$$

$$\Rightarrow\;\;sin\,\theta = \dfrac{v_{m/r}}{v_r}$$

Thus, the angle required to cross the river with minimum drift when speed of river is greater than speed of man is given by

$$\theta = sin ^{-1} \left(\dfrac{v_{m/r}}{v_r}\right)$$

A river is flowing at a speed of  $$\vec v_{river/Ground} = \; 5\; km/hr$$ . Tina can swim in still water with a speed  $$v_{Tina/river} = \; 3\;km/hr$$ . Calculate the angle  $$\theta$$ at which she should swim to cross the river with minimum drift.   $$\Big[sin\; 37 ° = \dfrac{3}{5}\Big]$$

A 60°

B 30°

C 45°

D 37°

×

The velocity of swimmer in still water is equal to the velocity of swimmer in reference frame attached to river.

Thus, $$\vec v_{Tina/river} = \; 3\;km/hr$$

and  $$\vec v_{river/Ground} = \; 5\; km/hr$$

For zero drift in horizontal direction, there should be no component of velocity of Tina in that direction.

$$\vec{v}_{T}=\vec{v}_{T/r}+\vec v_{r}$$

$$\vec v_{T} = (v_r-v_{T/r} \; sin\, \theta ) \hat i + v_{T/r} \; cos \,\theta \;\hat j$$

Drift, $$x = \vec v_{r/_G}\; × \; t_{min}\;\;\;\;[\because\;v_{r/_G} = v_r - v_{T/r}\; sin\, \theta]$$

$$x = (v_r-v_{T/r}\; sin\,\theta)\dfrac{d}{v_{T/r}\; cos \,\theta}$$

As,  $$|\vec v_r|>|\vec v _{T/r}|$$ thus, drift can't be zero. Because, $$sin \; \theta$$ always vary from –1 to 1.

When, $$sin \; \theta$$ is –1 then $$[v_r + v_{T/r}] \Rightarrow$$ drift is non-zero .

When,  $$sin \; \theta$$  is +1 then $$[v_r - v_{T/r}] \Rightarrow$$ drift is non-zero.

For minimum value of $$x,\; \dfrac{dx}{d\theta} \; = 0$$

$$\therefore\;\;\dfrac{d}{d \theta} \left(\dfrac{d\;v_r}{v_{T/r}\; cos \,\theta}\right) - \dfrac{d}{d\theta} \left[\dfrac{d\;v_{T/r }\;sin\, \theta}{v_{T/r}\; cos \,\theta}\right] = 0$$

$$\dfrac{dv_r}{v_{T/r}}\;\dfrac{d}{d\theta} (sec\; \theta) - d \left(\dfrac{d}{d\theta} tan\; \theta\right) = 0$$

$$d\left[\dfrac{v_r}{v_{T/r}}sec \,\theta \,tan\,\theta - sec^2\theta\right] = 0$$

$$(sec\,\theta)(d) \left[\dfrac{v_r}{v_{T/r}}\left(\dfrac{sin\,\theta}{cos\theta}\right) - \dfrac{1}{cos\, \theta}\right] = 0$$

$$\Rightarrow\;\;\dfrac{v_r\;\;sin\,\theta}{v_{T/r}\;\;cos\,\theta} = \dfrac{1}{cos\,\theta}$$

$$\Rightarrow\;\;sin\,\theta = \dfrac{v_{T/r}}{v_r}$$

Thus, the angle required to cross the river with minimum drift when speed of river is greater than speed of Tina is given by

$$\theta = sin ^{-1} \left(\dfrac{v_{T/r}}{v_r}\right)$$

For minimum drift,

$$\theta = sin^{-1} \left(\dfrac{\vec v_{Tina/river}}{\vec v_{river/Ground}}\right)$$

$$\theta = sin ^{-1} \left(\dfrac{3}{5}\right)$$

$$\theta = sin ^{-1} (sin\; 37 ° )$$

$$\theta = 37 °$$

A river is flowing at a speed of  $$\vec v_{river/Ground} = \; 5\; km/hr$$ . Tina can swim in still water with a speed  $$v_{Tina/river} = \; 3\;km/hr$$ . Calculate the angle  $$\theta$$ at which she should swim to cross the river with minimum drift.   $$\Big[sin\; 37 ° = \dfrac{3}{5}\Big]$$

A

60°

.

B

30°

C

45°

D

37°

Option D is Correct