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uh students we will discussing two three
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today so first question I would like to
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discuss today is uh the question States
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like that a particle moves in a straight
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relation x = tqb - 4 t² + 33 and we need
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to find the acceleration of particle
0:26
displacement equal to
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zero now in such typ of questions where
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we have a we have been given a function
0:34
we need to use of differential
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calculations to resolve such type of
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questions now first displacement is
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given us uh function time so first we
0:43
need to find the velocity of it and
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which can be find using as first
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differential of displacement so DX by DT
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is equal to 3 t² - 8 + 3 now the second
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differential will give the acceleration
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so again differentiating the equation we
1:02
get d2x S DT s = 6t - A so this is the
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acceleration of the particle at any
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point of time now here we need to find
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acceleration of those points where
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displacement is equal to zero so
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uh so it is clear we can find if we know
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the Val uh value of T now we need to
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find the value of T where the
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displacement is zero so uh putting xal
1:30
to0 in equation number one which is the
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equation solving this equation we get
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three values tal to 0 1 and 3 so at
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these three times the displacement of
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the particle is zero now we know the
1:44
acceleration equation so we can just
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simply put um ACC time in those
1:49
acceleration equation and we can find
1:54
now next question which I would like to
1:56
discuss is regarding the projectile
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so question is is like that which of the
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following remain constant during the
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motion of a projectile fired from a
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planet option given are kinetic energy
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momentum vertical comp velocity
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horizontal component velocity now to
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resolve this type of question we must
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know we must know the basic concept of
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the projectile what all goes along the
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way what all what all thing got changed
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so let's start with that the answer is D
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horizontal velocity of the component of
2:33
velocity now I will explain you how it
2:35
is like that see kinetic energy is given
2:38
as half MV Square now we know that
2:40
velocity changes in the projectile
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motion so ktic can have be constant
2:45
second moment again it depends on
2:47
velocity so it cannot be
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constant since now since acceleration
2:52
due to gravity acts during the motion
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velocity of the body changes throughout
2:55
the motion so kinetic energy and
3:02
now acceleration due to gravity acts in
3:04
vertical vertical appointment of
3:06
velocity change so here since the
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acceleration acting on the vertical
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Direction only so the vertical component
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of velocity is getting the horizontal
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compon of the Velocity is remaining
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constants so there's no horizontal
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acceleration horizontal component
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constant okay now the third question is
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a graphical question here we need to
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explain the here is a motion of a
3:32
particle along a straight line the
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velocity and time are given in this
3:36
graph we need to explain explation
3:39
motion okay in these type questions we
3:42
need to remember two things when the
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velocity time graph shows upward slope
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that means the body is accelerating when
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it show downward slope that means the
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body is retarding when the it's straight
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line with the time that mean velocity is
3:57
constant there's no acceleration
4:00
so here is like that so from time 0 to
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50 second the slope of the body is
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increasing slope of the body is remains
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acceleration remains constant from 1 to
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50 ACC the slope of the graph if you see
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the slope of the graph is decreasing
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so that mean if the slope of the graph
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is inre decreasing that means the
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acceleration value is decreasing the
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body is still accelerating but the
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acceleration is getting decreased and it
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becomes at this point it becomes
4:45
zero acceleration has become zero now
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going forward from 50 to 60 again now
4:53
body is retarding because the velocity
4:55
is decreasing but again the acceleration
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if you see the uh if you see it the
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acceleration in negative sense is
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getting increased that body is retarding
5:06
at some slow acceleration first and it's
5:08
slowly slowly increasing and it reaches
