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2003
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Kinematics
Scene 1
A pencil writes & draws the following (free hand):
Physics
Kinematics
(Sketch of a car, then the car speeds away)
Scene 2
On screen:
What you will see now is another IIT JEE problem on relative
velocity.
Voiceover:
What you will see now is another IIT JEE problem on relative
velocity.
Scene 3
On screen:
Example 1 (IIT JEE 2003)
A particle of mass m, moving in a circular path of radius R with
a constant speed2
v is located at
point (2R, 0) at time t = 0, and a man starts moving with a
velocity1v along the +ve y-axis from the
origin at time t = 0. Calculate the velocity of the particle
w.r.t the man as a function of time.
Voiceover:
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8/12/2019 Aristo Kinematics - Module 17 - Relative Vel - JEE
2003
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The problem is from IIT JEE 2003. It says - A particle of mass
m, moving in a circular path of radius R
with a constant speed2
v is located at point (2R, 0) at time t = 0, and a man starts
moving with a
velocity1v along the +ve y-axis from the origin at time t = 0.
Calculate the velocity of the particle
w.r.t the man as a function of time.
Scene 4
On screen:
So, if we write1
v and2
v in their vector form we will get 2 1v
1 1v v j
Voiceover:
We have been asked to find out the velocity of the particle
relative to the man, or 2 1v .
Now, 2 1 2 1v v v
So, if we write1
v and2
v in their vector form we will get 2 1v
It is very easy to write1
v in the vector form with respect to the coordinate axes
shown.
1 1v v j
Scene 5
On screen:
1: Man
2: Particle
We have to determine 2 1v
2 1 2 1v v v
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8/12/2019 Aristo Kinematics - Module 17 - Relative Vel - JEE
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( cos ) ( sin )r R R i R j
2v
R
2 ( sin ) ( cos ) ( sin ) ( cos )
d r d d v R i R j R i R j
dt dt dt
Or 2 2 2 ( sin ) ( cos )v v i v j
Voiceover:
Writing2
v is not that simple, though. That is because unlike1
v ,2
v is constantly changing in its
direction.
Suppose at time t, the particle is at point P. Then, itsposition
vector can be written as
( cos ) ( sin )r R R i R j
We must recognize here that r is a function of time because is
continuously changing with time.
Since the particle is moving in circle with constant speed 2v ,
its angular velocity is also constant.
2v
R
Lets now differentiate the position vector with respect to
time.
2 ( sin ) ( cos ) ( sin ) ( cos )
d r d d v R i R j R i R j
dt dt dt
Or 2 2 2 ( sin ) ( cos )v v i v j
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2003
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Scene 6
2 1 2 1v v v
But 1 1v v j
And 2 2 2 ( sin ) ( cos )v v i v j
So, 2 1 2 2 1 ( sin ) ( cos )v v i v v j
Voiceover:
We said earlier that
2 1 2 1v v v
And have found1 1
v v j
And 2 2 2 ( sin ) ( cos )v v i v j
So, 2 1 2 2 1 ( sin ) ( cos )v v i v v j
