Generally we consider the reference as earth
(i.e., stationary) and we will mention the velocities of all the bodies. So it
will be relative earth as earth velocity will be “O” its original velocity will
be equal to the relative velocity w.r.t earth.
But in the relative velocity concept
we will speak about the velocity of one body with respect to the other moving
body.
Relativity concept in scalar:
Let the age of A → 40 years
Let the age of B → 20 years
Then the age of “A” relative to B is
20.
Age of A - age of B i.e., 40 – 20 = 20
The age of “B” relative to A is “- 20”
Age of B - age of A = 20 - 40 = - 20
As the “age” is scalar we can straight
away subtract them but when we comes to vector for “ex”
Let the velocity of “A” be 20 m/s
along +ve x axis and the velocity of “B” be 20 m/sc along +ve y - axis. Then
relative velocity of A w.r.t B is VA - VB
Relative velocity of B w.r.t A is VB
- VA
⇒ From the parallel to green law
of vectors. We get,
|VA - VB| = √ (VA²
+ VB² + 2VA + VB cosθ)
θ is the angle between VA
and VB
⇒ |VA - VB|
= √ (VA² + VB² + 2VA + VB cos90)
⇒ |VA - VB|
= √ (VA² - VB²)
⇒ Direction of resultant = tan⁻¹
[(b sinθ)/ (a - b cosθ)]
(i) Relative motion of a boat in a river: Let vω be the velocity of water in a river flowing in
the direction RQ and vb be the velocity of the boar in still water
is shown in figure. The resultant velocity of the boat in the river is given by
v = √ (vω² + vb² + 2 vωvb cosθ)
a) To cross the river of width d along
the shortest path which is PQ, the boat must move along PR making an angle (900
+ θ) with the direction of the stream such that the direction of the resultant
velocity v is along PQ. Angle θ is given by sin θ = vω/vb
Also v = √ (vb² + vω²)
The time taken to cross the river
along the shortest path is given by t = d/v = d/ √ (vb² - vω²)
b) To cross the river in the shortest
time, the boat should move along PQ. The shortest time is given by t = d/vb.
At this time, the boat will reach the point R on the opposite bank of the river
at a distance x from the point Q. From the Figure, we have x = d tan θ, but tan
θ = vω/vb. Therefore, x = d (vω/vb)
(ii) Holding an Umbrella to Protect from Rain: let vr be the velocity of the rain falling vertically
downward and vm the velocity of a man walking from north to south
direction. In order to protect himself from rain, the must hold his umbrella in
the direction of the resultant velocity v, which is given by v = √ (vr²
+ vm²).
This is the speed with which the rain strikes the umbrella. If θ is the angle
subtended by the resultant velocity v with the vertical, then from triangle
ORM`, we have tan θ = RM’/ OR = vm/vr or θ = tan⁻¹ (vm
/ vr).
Thus the man must hold the umbrella at
an angle θ with the vertical towards south.
Example: A jet airplane travelling at the speed of 500 km h⁻¹
ejects its products of combustion at the speed of 1500 km h⁻¹ relative to the
jet plane. What is the speed of the latter with respect to an observer on the
ground?
Solution:
1500î = Vcg +500î
No comments:
Post a Comment