### Physical Significance of Relative Velocity

Two persons A & B are in the two vehicles moving in the same direction as shown in the figure.

Assume, v_{A} = 10m/sec & v_{B} = 4m/sec

The person A notices the person B moving towards him with a speed of (10 − 4) m/s = 6 m/sec.

This is the velocity of B with respect to ( or relative to) A . $ \displaystyle \vec{v_{BA}} $ is directed from B to A.

Similarly A seems to move towards B with a speed of 6 m/sec.

Therefore the velocity of A relative to B ( $ \displaystyle \vec{v_{AB}} $) has a magnitude of 6 m/sec & is directed from A to B as shown in the figure.

Therefore $ \displaystyle \vec{v_{AB}} = – \vec{v_{BA}}$

In general, $ \displaystyle \vec{v_{BA}} = \vec{v_B}-\vec{v_A} $

$ \displaystyle |\vec{v_{BA}}| = |\vec{v_{AB}}| $

$ \displaystyle v_{AB} = \sqrt{v_A^2 + v_B^2 – 2 v_A v_B cos\theta} $

Solved Example : A train moves due east with a velocity v_{1} = 20m/sec and a car moves due north with a velocity v_{2} = 15 m/sec. Find the velocity of the car as observed by a passenger sitting in the train.

Solution : The passenger observes the velocity of the car w.r.t. himself.

That is, the velocity of the car relative to the train is

$ \displaystyle \vec{v_{ct}} = \vec{v_c}-\vec{v_t} $

$ \displaystyle \vec{v_{21}} = \vec{v_2}-\vec{v_1} $

Magnitude of v_{21}^{→} is

$ \displaystyle v_{21} = \sqrt{v_1^2 + v_2^2 } $

$ \displaystyle v_{21} = \sqrt{(20)^2 + (15)^2 } $

= 25 m/s

Direction of v_{21} :

$ \displaystyle \phi = tan^{-1}\frac{20}{15}$

$ \displaystyle \phi = tan^{-1}\frac{4}{3}$ , west of north

### Relative Motion between Rain & Man

We know that

v_{r} ≡ v_{rg} = velocity of rain w.r.t. ground

v_{m} ≡ v_{mg} = velocity of man w.r.t. ground

$ \displaystyle \vec{v_{rm}} = \vec{v_r}-\vec{v_m} $

$ \displaystyle \vec{v_r} = \vec{v_{rm}} + \vec{v_m} $

That means the vector addition of the velocity of rain with respect to man ( v_{rm}^{→}) and the velocity of man (vehicle) yields the actual velocity of rain v_{r}^{→} .

The magnitude and direction of v_{r}^{→} can be given as

$ \displaystyle v_r = \sqrt{v_{rm}^2 + v_m^2 + 2 v_{rm} v_m cos\theta} $

$ \displaystyle tan\phi = \frac{v_{rm} sin\theta}{v_m + v_{rm} cos\theta} $ , with horizontal ( v_{m}^{→} )

Solved Example : A man holds his umbrella vertically up while walking due west with a constant velocity of magnitude v_{m} = 1.5 m/sec in rain. To protect himself from rain, he has to rotate his umbrella through an angle φ = 30° when he stops walking. Find the velocity of the rain.

Solution :

$ \displaystyle \vec{v_r} = \vec{v_{rm}} + \vec{v_m} $

Since , v_{rm}^{→} ⊥ v_{m}^{→}

we obtain ,

$ \displaystyle sin\phi = \frac{v_m}{v_r} $

⇒ v_{r} = v_{m} cosec φ

⇒ v_{r} = 1.5 × cosec 30°

= 1.5 ×2 = 3m/sec .

Exercise : A boy is running on a horizontal road with a velocity of 5 m/s. At what angle should he hold his umbrella in order to protect himself from the rain, if it is raining with a velocity of 10 m/s vertically downward ?

### Relative Motion of Swimmer in Flowing Water

Since , $ \large \vec{V_{mw}} = \vec{V_m} -\vec{V_w}$

$ \large \vec{V_m} = \vec{V_{mw}} + \vec{V_w}$

Illustration : A man swims at an angle θ = 120° to the direction of water flow with a speed V_{mw} = 5km/hr relative to water. If the speed of water V_{w} = 3km/hr, find the speed of the man.

Sol:

$ \displaystyle \vec{V_{mw}} = \vec{V_m} -\vec{V_w}$

$ \displaystyle \vec{V_m} = \vec{V_{mw}} + \vec{V_w}$

$ \displaystyle V_m = \sqrt{V_{mw}^2 + V_w^2 + 2 V_{mw} V_w cos\theta}$

$ \displaystyle V_m = \sqrt{5^2 + 3^2 + 2(5)(3) cos120} $

$ \displaystyle V_m = \sqrt{25 + 9 -15} $

$ \displaystyle V_m = \sqrt{19} m/s $

Exercise : To a man walking at 7 km/ hr due west the wind appears to blow from north west, but when he walks at 3 km/hr due west the wind appears to blow from the north. What is the actual direction of the wind & what is its velocity ?

### Crossing of the river with Minimum drift

CASE -1 : V_{mw} > V_{w}

A man intends to reach the opposite bank at the point directly opposite to the stationary point. He has to swim at angle θ with a given speed V_{mw} w.r.t. water, such that his actual velocity Vm will direct along AB, that is perpendicular to the bank (or velocity of water V_{w} )

For minimum drift ,

$ \displaystyle \vec{V_m} \perp \vec{V_w} $

You can realise the situation by a simple example. If you want to reach the directly opposite point or cross the river perpendicularly, a man, that is to say, Hari, must report you that, you are moving perpendicular to the shore. What does this report signify? Since Hari observes your actual velocity (V_{m} ) to be perpendicular to the bank is perpendicular to V_{w} .

Observing the vector-triangle V_{w} = V_{mw} sinθ & V_{m} = V_{mw} cosθ

$ \displaystyle \theta = sin^{-1}\frac{V_w}{V_{mw}} $

$ \displaystyle V_m = \sqrt{V_{mw}^2 – V_w^2} $

The time of crossing

$ \displaystyle t = \frac{d}{V_m} = \frac{d}{\sqrt{V_{mw}^2 – V_w^2}}$

Solved example : The speed of a man in a pond is double that of the water in a river. The man starts swimming from a point P on the bank. What is the angle at which the man should swim so as to get directly to the opposite bank.

Solution : Let the speed of the man w.r.t. water be Vmw

When the man swims in a river (moving water), the velocity of the man is directly across the river, i.e.

$ \displaystyle \vec{V_m} \perp \vec{V_w} $ ;

if the swims at an angle φ w.r.t. water,

$ \displaystyle sin\phi = \frac{V_w}{V_{mw}} $ …(1)

Since the speed of water is half of that of the man relative to water (speed of the man in still water)

V_{mw} = 2 V_{w}

$ \displaystyle \frac{V_w}{V_{mw}} = \frac{1}{2} $ …(ii)

From (i) & (ii

sinφ = 1/2 ⇒ φ = 30°

The angle of swimming = θ = 90 +φ= 90° + 30° = 120° to the direction of flow of water.

Exercise : Referring to the previous illustration , find the velocity of the swimming man, if speed of water is 3 km/hr.

CASE -2 : V_{mw} < V_{w}

Let the man head at an angle θ with normal to the bank for minimum drift. Suppose the drift is equal to zero. For zero drift, the velocity of the man along the bank is zero.

V_{m} = V_{w} – V_{mw} sin θ = 0

This gives,

$ \displaystyle sin\theta = \frac{V_w}{V_{mw}} $

Since , Vw > Vmw ;

sin θ > 1 ; which is impossible. Therefore the drift can not be zero.

Now let the man head at an angle θ with normal to the bank to experience minimum drift. Suppose that the drifting of the man during time t when he reaches the opposite bank is

BC = x.

x = (V_{mx})t …(i)

(V_{mx}) = component of velocity of man along the water flow

Where $ \displaystyle t = \frac{d}{V_{mw} cos\theta} $ …(ii)

(V_{mx}) = V_{w} – V_{mw} sinθ. …(iii)

Using (i), (ii) & (iii) we obtain

$\large x = (V_w – V_{mw}sin\theta) \frac{d}{V_{mw} cos\theta} $

$\large x = (\frac{V_w}{V_{mw}} sec\theta – tan\theta )d $ …(iv)

For x to be minimum, $\frac{dx}{d\theta} = 0 $

$\large (\frac{V_w}{V_{mw}}sec\theta tan\theta – sec^2\theta ) d = 0 $

$\large \frac{V_w}{V_{mw}}tan\theta = sec\theta $

$\large sin\theta = \frac{V_{mw}}{V_w} $

Substituting the value of θ in (iv) we obtain,

$\large x = ( \frac{\sqrt{V_w^2 – V_{mw}^2}}{V_{mw}} ) d $

Solved Example : The speed of water is double that of (swimmer) relative to water. What should be the direction of the swimmer so as to experience minimum displacement assuming V_{m} = 5 m/sec ?

Solution: As , $\large sin\theta = \frac{V_{mw}}{V_w} $

Putting the value of $\large \frac{V_{mw}}{V_w} = \frac{1}{2} $ in the derived expression

$\large sin\theta = \frac{1}{2} $

θ = 30°

Exercise : Referring to previous illustration, What is the time of crossing of the man assuming width of the river is 500 m ?

### Crossing of the River in Minimum Possible Time

Case 1: To reach the opposite bank for a given v_{mw}

Let the man swim at an angle θ with AB. We know that the component of the velocity of man along shore is not responsible for its crossing the river. Only the component of velocity of man (V_{m}) along AB is responsible for its crossing along AB.

The time of crossing $\large t = \frac{AB}{V_{mw} cos\theta }$

Time is minimum when cos θ is maximum

The maximum value of cos θ is 1 for θ = 0

That means the man should swim perpendicular to the shore

$\large \vec{V_{mw}} \perp \vec{V_w} $

$\large t_{min} = \frac{d}{V_{mw} cos\theta}$

At θ = 0 ;

$\large t_{min} = \frac{d}{V_{mw}}$

Solved Example : A man crosses the river in shortest time at an angle θ = 60° to the direction of flow of water. If the speed of water is v_{w} = 4 km/hr, find the speed of the man.

Solution : Referring to the theory, we know that for minimum time of crossing the man should head perpendicular to the shore

$\large \vec{V_{mw}} \perp \vec{V_w} $

Therefore , $\large cos\theta = \frac{V_w}{V_m}$

$\large cos60^o = \frac{4}{V_m}$

V_{m} = 8 km/h

Exercise : Referring to previous illustration,

(a) If the width of the river is (1/2) km, find the displacement of the man in crossing the river.

(b) What is the drift of water on the man ?

Case 2 : To reach directly opposite point on the other bank for a given v_{mw} & velocity V of walking along the shore.

To attain the direct opposite point B in the minimum time. Let the man swim at an angle θ with the direction AB.

The total time of journey t = the time taken from A to C + the time taken from C to B

$\large t = t_{AC} + t_{CB}$

Where , $\large t_{AC} = \frac{AB}{V_{mw}cos\theta } $ & $\large t_{CB} = \frac{BC}{V} $

Where V = walking speed of the man from C to B.

$\large t = \frac{AB}{V_{mw}cos\theta} + \frac{BC}{V} $ ….(i)

Again BC = (V_{m})_{x} t

$\large BC = (V_w – V_{mw} sin\theta ) (\frac{AB}{V_{mw}cos\theta})$

From (i)

$\large t = \frac{AB}{V_{mw}cos\theta} + (\frac{(V_w – V_{mw} sin\theta )AB}{V (V_{mw}cos\theta}) $

$\large t = \frac{d}{V_{mw}} [(1 +\frac{V_m}{V})\frac{sec\theta}{V_{mw}} – \frac{tan\theta}{V}] $

Putting $\frac{dt}{d\theta} = 0 $ for minimum t we obtain

$\large \frac{dt}{d\theta} = \frac{d}{d\theta}(\frac{d}{V_{mw}} [(1+\frac{V_m}{v})\frac{sec\theta}{V_{mw}} – \frac{tan\theta}{V}]) = 0 $

$\large \frac{sec\theta tan\theta}{V_{mw}}(1+\frac{V_m}{V}) – \frac{sec^2 \theta}{V} = 0 $

$\large \frac{tan\theta}{V_{mw}}(1 +\frac{V_m}{V}) = \frac{sec\theta}{V} $

$\large sin\theta = \frac{V_{mw}}{V + V_w} $

This expression is obviously true when V_{mw} < V + V_{w}

Solved Example : A man can walk on the shore at a speed V_{1} = 6 km/hr & swim in still water is a speed V_{2} = 5 km/hr. If the speed of water is V_{3} = 4 km/hr, at what angle should he head in the river in order to reach the right opposite shore in shortest time including his swimming & walking ?

Solution : Directly using the previous result we obtain the angle of swimming

$\large sin\theta = \frac{V_{mw}}{V + V_w} $

Putting V = V_{1} = 6 km/hr, V_{mw} = V_{2} = 5 km/hr

V_{w} = V_{3} = 4 km/hr , we obtain ,

$\large sin\theta = \frac{1}{2} $

⇒ θ = 30°

The man should head to an angle , α = 90 + θ = 120° with the direction of flow of water.

Exercise : Referring to previous illustration, What is the time elapsed for swimming, if the width of the river is (1/2) km?