## Relative Velocity Of Rain With Respect To A Moving Observer

In the absence of wind, rain falls vertically. But when a man moves forward during a rainfall, he has to slant his umbrella in front of him. This is due to the fact that the relative velocity of the falling rain with respect to the man makes an angle with the vertical.

Let the actual downward velocity of rain = \(\vec{v}\); the horizontal velocity of the man = \(\vec{u}\).

Hence, the relative velocity of rain with respect to the man, \(\vec{w}\) = \(\vec{v}\) – \(\vec{u}\)

∴ w = \(\sqrt{v^2+u^2}\)….(1)

If the relative velocity makes an angle θ with the vertical, then, tanθ = 2……(2)

From equation (1) the magnitude of the relative velocity of rain and from equation (2) its direction can be determined. As the relative velocity of rain makes an angle θ with the vertical, it seems that rain comes down at an angle when we move. This is why, while riding a cycle, one has to hold an umbrella along CO to save himself from getting wet.

For the same reason, a speeding vehicle receives more rain on the front windscreen than on the rear one. Similarly, a person sitting inside a moving train feels that the raindrops follow a slanted path.

**Read and Learn More: Class 11 Physics Notes**

## Relative Velocity Of Rain With Respect To A Moving Observer Numerical Examples

**Example 1. A man is walking on a horizontal road at 3 km · h ^{-1} t while rain is falling vertically with a velocity of 4 km · h^{-1}. Find the magnitude and direction of the i velocity of rain with respect to the man.**

**Solution:**

Velocity of the man, u = 3 km· h^{-1}; velocity of rainfall, v = 4 km · h^{-1}

Since, ∠AOB = 90°, the magnitude of the relative velocity of rain with respect to the man,

w = \(\sqrt{u^2+v^2}=\sqrt{3^2+4^2}=5 \mathrm{~km} \cdot \mathrm{h}^{-1}\)

If \(\angle B O C=\theta, \tan \theta=\frac{u}{v}\) or, \(\theta=\tan ^{-1} \frac{u}{v}\)

∴ \(\theta=\tan ^{-1} \frac{3}{4}=36.9^{\circ}\)

Hence, the velocity of rain with respect to the man makes an angle of 36.9° with the vertical.

**Example 2. To a man, walking on a horizontal path at 2 km · h ^{-1}, rain appears to fall vertically at 2 km · h^{-1}. Find the magnitude and the direction of the actual velocity of rainfall.**

**Solution:**

⇒ \(\vec{u}\) = velocity of the man, \(\vec{v}\)= actual velocity of the raindrops ; so, the relative velocity of the rain¬drops with respect to the man

⇒ \(\vec{w}=\vec{v}-\vec{u} \quad \text { or, } \vec{v}=\vec{u}+\vec{w}\)

Therefore, the resultant of the velocity of the man (\(\vec{u}\)) and the relative velocity of rain (\(\vec{w}\)) will give the real velocity of rain (\(\vec{v}\)).

From the figure, \(v=\sqrt{2^2+2^2}=2 \sqrt{2} \mathrm{~km} \cdot \mathrm{h}^{-1}\)

If the actual direction of rain is inclined at an angle θ with the vertical, then

tanθ = \(\frac{u}{w}\) = \(\frac{2}{2}\) = 1

∴ θ = 45°.

**Example 3. To a car driver moving at 40 km · h ^{-1} towards the south, the wind appears to blow towards the east. When the speed of the car is reduced to 20km· h^{-1} wind appears to blow from the north-west. Find the magnitude and direction of the actual velocity of the wind.**

**Solution:**

Let us choose: the x-axis along the east and the y-axis along the north.

Initial velocity of the car, \(\vec{u}_1=-40 \hat{j} \mathrm{~km} \cdot \mathrm{h}^{-1}\)

Final velocity of the car, \(\vec{u}_2=-20 \hat{j} \mathrm{~km} \cdot \mathrm{h}^{-1}\)

Let, \(\vec{v}\) actual velocity of the wind.

Then, the relative velocity of the wind with respect to the car, \(\vec{w}=\vec{v}-\vec{u} \quad \text { or, } \vec{v}=\vec{u}+\vec{w}\)

Initially, \(\vec{w}_1=w_1 \hat{i}\); so \(\vec{v}=\vec{u}_1+\vec{w}_1=w_1 \hat{i}-40 \hat{j}\)…..(1)

Finally, \(\vec{w}_2=w_2 \cos 45^{\circ} \hat{i}-w_2 \sin 45^{\circ} \hat{j}\) (as it is towards south-east)

= \(\frac{w_2}{\sqrt{2}} \hat{i}-\frac{w_2}{\sqrt{2}} \hat{j}\)

∴ \(\vec{v}=\vec{u}_2+\vec{w}_2=\frac{w_2}{\sqrt{2}} \hat{i}-\left(\frac{w_2}{\sqrt{2}}+20\right) \hat{j}\)

Comparing the coefficients of \(\hat{j}\) in (1) and (2), \(\frac{w_2}{\sqrt{2}}+20=40 \text { or, } \frac{w_2}{\sqrt{2}}=20\)

Then from (2), \(\vec{v}=20 \hat{i}-(20+20) \hat{j}=(20 \hat{i}-40 \hat{j}) \mathrm{km} \cdot \mathrm{h}^{-1}\) (between east and south)

Magnitude of \(\vec{v}=v=\sqrt{(20)^2+(40)^2}=20 \sqrt{1+4}\)

= \(20 \sqrt{5} \mathrm{~km} \cdot \mathrm{h}^{-1}\)

If \(\vec{v}\) is inclined at an angle θ with east, then \(\tan \theta=\frac{-40}{20}=-2=\tan \left(-63.4^{\circ}\right) \quad \text { or, } \theta=-63.4^{\circ}\)

So, the wind velocity is at an angle of \(63.4^{\circ}\) south of east.

## Application Of Relative Velocity In One An Two Dimensional Motion

**Motion Of A Boat On A River**

Let \(\vec{v}\) be the speed of a boat in still water. If the river current has a velocity of u along its length, the boat is subjected to this velocity too. Therefore, it will move with a resultant velocity, \(\vec{w}\) = \(\vec{v}\) + \(\vec{u}\), and the motion of the boat will, in general, be oblique.

**Shortest Distance Traversed In Crossing The River:** To find the direction of \(\vec{v}\) such that the distance traveled by the boat is the least, let the angle made by \(\vec{v}\) with the shortest distance of crossing be θ. Resolving the components of \(\vec{v}\) in two mutually perpendicular directions

- Against the current \(\vec{u}\) and
- In the direction of the shortest distance AB, we get v sinθ and v cosθ respectively. The component v sinθ balances the effect of the current u.

Therefore, \(u=v \sin \theta \quad \text { or, } \theta=\sin ^{-1} \frac{u}{v}\)

Component v cosθ enables the boat to cross the river along AB and the time t required for that is

∴ t = \(\frac{l}{v \cos \theta}\)……(1)

The vector vcosθ is actually \(\vec{v}\) + \(\vec{u}\) = \(\vec{w}\) = the resultant velocity. So, to cross along the shortest distance, the boat should be steered in such a way that the resultant velocity is directly across the river.

w = \(\sqrt{v^2-u^2}\), and the angle of steering is, \(\theta=\sin ^{-1} \frac{u}{v}\). If l be the width of the river, the time taken to cross it is,

t = \(\frac{l}{w}=\frac{l}{\sqrt{v^2-u^2}}\)….(2)

**Crossing The River In Minimum Time:** In equation (1). t will be minimum when the value of cos# is maximum, i.e., cosθ = 1 or, θ = 0°. Hence, to cross the river in minimum time, the boat should be driven along the width of the river.

In that case, the situation is as shown. The boat crosses obliquely along AB with the resultant velocity w. The time required to cross the river is

t’ = \(\frac{A B}{w}=\frac{A C}{v}=\frac{l}{v}\)…..(3)

This is exactly the time required when no current is present. The longitudinal displacement of the boat due to oblique crossing is, \(C B=A C \tan \alpha=\frac{l u}{v}\)….(4)

If there is no current, u=0. Then from equations (2), (3), and (4), we get, t = \(t^{\prime}=\frac{l}{v} \quad \text { and } C B=0\)

## Motion Of A Boat On A River Numerical Examples

**Example 1. The velocity of a boat in still water is 5 km · h ^{-1}. It takes 15 min to cross a river along the width. The river is 1 km wide. Find the velocity of the current.**

**Solution:**

The boat crosses a river of width 1 km in 15 min or in \(\frac{1}{4}\) h.

The resultant of the velocity of the boat and that of the current is \(\vec{w}\)(say) and

∴ w = \(\frac{1}{1 / 4}=4 \mathrm{~km} \cdot \mathrm{h}^{-1} \text {. }\)

Let the velocity of the boat in still water be \(\vec{v}\) and the velocity of the current be \(\vec{u}\).

Frome figure, \(v^2=u^2+w^2\)

or, \(u^2=v^2-w^2\)

or, \(u=\sqrt{v^2-w^2}=\sqrt{5^2-4^2}=3 \mathrm{~km} \cdot \mathrm{h}^{-1}\)

**Example 2. A man can reach the point directly opposite on the other bank of a river by swimming across the river in time t _{1} and crossing the same distance in time t_{2} while swimming along the current. If the velocity of the man in still water is v and the velocity of the water current is u, find the ratio between t_{1} and t_{2}.**

**Solution:**

Let the resultant velocity with which the man swims across the river be w. Hence, w = \(\sqrt{v^2-u^2}\) and therefore, time required to cross the river,

∴ \(t_1=\frac{l}{\sqrt{v^2-u^2}}\)…..(1)

Where l is the width of the river

When the man swims in the direction of the current, the resultant velocity, w’ = v+u, and time required to cross the same distance,

∴ \(t_2=\frac{l}{v+u}\)…….(2)

From (1) and (2), \(\frac{t_1}{t_2}=\frac{v+u}{\sqrt{v^2-u^2}}=\frac{v+u}{\sqrt{v-u} \cdot \sqrt{v+u}}=\sqrt{\frac{v+u}{v-u}}\)

**Example 3. Two boats, each with a velocity 8 km · h ^{-1}, attempt to cross a river of width 800 m. The velocity of river current is 5 km · h^{-1}. One of the boats crosses the river following the shortest path and the other follows the route in which the time taken is minimum. If they start simultaneously, what would be the time difference between their arrivals at the other bank?**

**Solution:**

In order to follow the shortest path, the boat should set itself at a particular angle with the current such that the boat’s resultant velocity is perpendicular to the direction of the current. Hence, the magnitude of the resultant velocity of the boat,

w = \(\sqrt{v^2-u^2}\)

and the time required to cross the river along the shortest path, \(t_1=\frac{x}{w}=\frac{x}{\sqrt{v^2-u^2}}=\frac{0.8}{\sqrt{8^2-5^2}}=0.128 \mathrm{~h}=7.69 \mathrm{~min}\)

[x = 800 m = 0.8 km, v = 8 km · h^{-1} and u = 5 km · h^{-1}]

To cross in minimum time, the boat should travel at right angles to the current and the time required is

Required time difference, t_{1} – t_{2} = 7.69 – 6 = 1.69 min

**Example 4. A person can swim at 4 km · h ^{-1} in still water. At what angle should he set himself to cross the river in a direction perpendicular to the river current of velocity 2 km · h^{-1}?**

**Solution:**

Let the velocity of the person be \(\vec{v}\), velocity of the current be \(\vec{u}\) and the resultant of \(\vec{u}\) and \(\vec{v}\) be \(\vec{w}\). The person crosses the river along its width from one bank to the other.

Hence, \(\vec{w}\) and \(\vec{u}\) are perpendicular to each other. If the angle between \(\vec{v}\) and \(\vec{w}\) is θ, then \(\sin \theta=\frac{u}{v}\)= \(\frac{2}{4}=\frac{1}{2}\) or, \(\theta=30^{\circ} \text {. }\)

Hence, the person has to swim at an inclination of 30° with the width of the river, or at an angle of (90° + 30°) or, 120° with the direction of the current as shown.

**Example 5. The width of a river is D. A man can cross the river in time t _{1} in the absence of any river current. But in the presence of a certain river current, the man takes a time t_{2} to cross the river directly. Show that the velocity of the current is v = \(D \sqrt{\frac{1}{t_1^2}}-{\frac{1}{t_2^2}}\)**

**Solution:**

Let u = velocity of the man on the river. \(t_1=\frac{D}{u} \quad \text { or, } u=\frac{D}{t_1}\)….(1)

In the presence of a river current of velocity v, the effective velocity of the man will be the resultant w of u and v. \(\vec{w} = \vec{u}+\vec{v}\)

To cross the river directly, the man has to swim in such a direction that the resultant w is across the width of the river.

Clearly, w = \(\sqrt{u^2-v^2}\)

and the time taken to cross the river is \(t_2=\frac{D}{w}=\frac{D}{\sqrt{u^2-v^2}}\).

∴ \(t_2^2=\frac{D^2}{u^2-v^2}\) or, \(u^2=v^2+\frac{D^2}{t_2^2}\)……(2)

From (1) and (2), \(\frac{D^2}{t_1^2}=v^2+\frac{D^2}{t_2^2}\) or, \(v^2=D^2\left(\frac{1}{t_1^2}-\frac{1}{t_2^2}\right)\)

∴ v = \(D \sqrt{\frac{1}{t_1^2-\frac{1}{t_2}}}\)