Relative Velocity Explanation , Solved Examples

Physical Significance of Relative Velocity

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

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Assume, vA = 10m/sec & vB = 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}}$

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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 v1 = 20m/sec  and a car moves due north with a velocity v2 = 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

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

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

Magnitude of v21 is

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

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

= 25 m/s

Direction of v21 :

$ \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

vr ≡ vrg = velocity of rain w.r.t. ground

vm ≡ vmg = 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 ( vrm) and the velocity of man (vehicle) yields the actual velocity of rain vr .

The magnitude and direction of vr 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 ( vm )

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Solved Example  : A man holds his umbrella vertically up while walking due west with a constant velocity of magnitude vm = 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 ,  vrm ⊥ vm

we obtain ,

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

⇒ vr = vm cosec φ

⇒ vr = 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}$

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

Sol:

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$ \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 : Vmw > Vw

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 Vmw w.r.t. water, such that his actual velocity Vm will direct along AB, that is perpendicular to the bank (or velocity of water Vw )

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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 (Vm ) to be perpendicular to the bank is perpendicular to Vw .

Observing the vector-triangle Vw = Vmw sinθ   &  Vm = Vmw 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.

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$ \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)

Vmw = 2 Vw

$ \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 : Vmw <  Vw

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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.

Vm = Vw – Vmw 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 = (Vmx)t …(i)

(Vmx) = component of velocity of man along the water flow

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

(Vmx) = Vw – Vmw 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 Vm = 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 vmw

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 (Vm) along AB is responsible for its crossing along AB.

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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 vw = 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

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

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

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

Vm = 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 vmw & 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

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

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 = (Vm)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 Vmw < V + Vw

Solved Example : A man can walk on the shore at a speed V1 = 6 km/hr & swim in still water is a speed V2 = 5 km/hr. If the speed of water is V3 = 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 = V1 = 6 km/hr, Vmw = V2 = 5 km/hr

Vw = V3 = 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?

Also Read :

→ Rest & Motion relative to Frame of reference
→ Average Speed & Average Velocity
→ Average Acceleration , Instantaneous Acceleration , Average Speed & Average Velocity
→ Uniformly & Non-Uniformly Accelerated Motion
→ Vertical motion under gravity
→ Relative Velocity

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