Contents
Physics Topics cover a broad range of concepts that are essential to understanding the natural world.
What is the Application of Relative Velocity?
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 = \(\vec{u}\) [Fig.].
Hence, 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 = \(\frac{u}{v}\) ….. (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 O 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 wind-screen than on the rear one. Similarly, a person sitting inside a moving train feels that the rain drops follow a slanted path.
Numerical Examples
Example 1.
A man is walking on a horizontal road at 3 km ᐧ h-1 while rain is falling vertically with a velocity of 4 km ᐧ h-1 . Find the magnitude and direction of the velocity of rain with respect to the man. [HS ‘04]
Solution:
Velocity of the man, u = 3 km ᐧ h-1; velocity of rainfall, u = 4 km ᐧ h-1 [Fig.]
Since, ∠AOB = 90°, the magnitude of relative velocity of rain with respect to the man,
w = \(\sqrt{u^2+v^2}\) = \(\sqrt{3^2+4^2}\) = 5 km ᐧ h-1
If ∠BOC = θ, tanθ = \(\frac{u}{v}\) or, θ = tan-1\(\frac{u}{v}\)
∴ θ = tan-1\(\frac{3}{4}\) = 36.9°
Hence, the velocity of rain with respect to the man makes an angle 36.9° with the vertical.
Example 2.
To a man, walldng on a horizontal path at 2 km ᐧ h-1, rain appears to fall vertically at 2 km ᐧ h-1. Find the magnitude and the direction of the actual velocity of rainfall. [HS ‘04]
Solution:
In Fig., \(\vec{u}\) = velocity of the man, \(\vec{v}\) = actual velocity of the raindrops; so, the relative velocity of the rain drops with respect to the man
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 ram (\(\vec{v}\)).
From the figure, v = \(\sqrt{2^2+2^2}\) = 2\(\sqrt{2}\) km ᐧ 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 ᐧ h-1 towards south, wind appears to blow towards east. When the speed of the car is reduced to 20 km ᐧ h-1 wind appears to blow from north-west. Find the magnitude and direction of the actual velocity of the wind.
Solution:
Let us choose: x-axis along east and y-axis along north.
Initial velocity of the car, \(\vec{u}_1\) = -40\(\hat{j}\) km ᐧ h-1
Final velocity of the car, \(\vec{u}_2\) = -20\(\hat{j}\) km. hᐧ h-1
Let, \(\vec{v}\) = actual velocity of the wind
Then, relative velocity of the wind with respect to the car,
If \(\vec{v}\) is inclined at angle θ with east, then
tanθ = \(\frac{-40}{20}\) = -2 = tan(-63.4°) or θ = -63.4°
So, the wind velocity is at an angle of 63.4° south of east.
Application Of Relative Velocity In One And Two Dimentional 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 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 travelled 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 in two mutually perpendicular directions-
- against the current \(\vec{u}\) and
- in the direction of shortest distance AB [Fig.], we get v sinθ and v cosθ respectively. The component vsinθ balances the effect of the current u.
Therefore, u = v sinθ or, θ = sin-1\(\frac{u}{v}\)
Component vcosθ enables the boat to cross the river along AB and the time t required for that is
∴ t = \(\frac{l}{v \cos \theta}\) …. (1)
In Fig., 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. From the Fig.,
w = \(\sqrt{v^2-u^2}\), and the angle of streering is, θ = 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 cosθ 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 in Fig. 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 Fig. when no current is present. The longitudinal displacement of the boat due to oblique crossing is,
CB = ACtanα = \(\frac{l u}{v}\) …… (4)
If there is no current, u = 0. Then from equation (2), (3) and (4), we get,
t = t’ = \(\frac{l}{v}\) and CB = 0
Numerical Examples
Example 1.
Velocity of a boat in still water is 5 km ᐧ h-1. It takes 15 min to cross a river along the width. The river is 1 km wide. Find the velocity of 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 km ᐧ h-1
Let the velocity of the boat in still water be \(\vec{v}\) and the velocity of the current be \(\vec{u}\). From Fig.,
v2 = u2 + w2
or, u2 = v2 – w2
or, u = \(\sqrt{v^2-w^2}\) = \(\sqrt{5^2-4^2}\) = 3 km ᐧ 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 t1 and crosses the same distance in time t2 while swimming along the current. If the velocity of the man in still water is v and velocity of the water current is u, find the ratio between t1 and t2.
Solution:
Let the resultant velocity with which the man swims across the river be w. Hence, w = \(\sqrt{v^2-u^2}\) [Fig.] and therefore, time required to cross the river,
t1 = \(\frac{l}{\sqrt{v^2-u^2}}\), …… (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,
t2 = \(\frac{l}{v+u}\) ….. (2)
From (1) and (2),
\(\frac{t_1}{t_2}\) = \(\frac{v+u}{\sqrt{v^2-u^2}}\) = \(\frac{v+u}{\sqrt{v-u} \cdot \sqrt{\nu+u}}=\) = \(\sqrt{\frac{v+u}{v-u}}\)
Example 3.
Two boats, each with velocity 8 km ᐧ h-1, attempt to cross a river of width 800 m. The velocity of river current is 5 km ᐧ 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 [Fig.(a)]. Hence, the magnitude of the resultant velocity of the boat,
and the time required to cross the river along the shortest path,
t1 = \(\frac{x}{w}\) = \(\frac{x}{\sqrt{v^2-u^2}}\) = \(\frac{0.8}{\sqrt{8^2-5^2}}\) = 0.128 h = 7.69 min
[x = 800m = 0.8 km, v = 8 km ᐧ h-1 and u = 5km ᐧ h-1]
To cross in minimum time, the boat should travel at right angles to the current [Fig.(b)] and the time required is
t2 = \(\frac{x}{v}\) = \(\frac{0.8}{8}\) = 0.1 h = 6 min
∴ Required time difference,
t1 – t2 = 7.69 – 6 = 1.69 min
Example 4.
A person can swim at 4 km ᐧ 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 ᐧ h-1. [HS(XI) ‘06]
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 \(\vec{w}\) [Fig.].
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 from Fig.,
sinθ = \(\frac{u}{v}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\) or, θ = 30°.
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 in Fig.
Example 5.
The width of a river is D. A man can cross the river in time t1 in the absence of any river current. But in presence of a certain river current, the man takes a time to cross the river directly. Show that the velocity of the current is
v = \(D \sqrt{\frac{1}{t_1^2}-\frac{1}{t_2}}\)
Solution:
Let u = velocity of the man on the river.
∴ t1 = \(\frac{D}{u}\) or, u = \(\frac{D}{t_1}\) …… (1)
In presence of 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 [Fig.]. Clearly,
w = \(\sqrt{u^2-v^2}\)