Civil Engineering (CE) Exam > Civil Engineering (CE) Videos > Kinematics Velocity and Acceleration
Kinematics Velocity and Acceleration Video Lecture - Civil Engineering (CE)
FAQs on Kinematics Velocity and Acceleration Video Lecture - Civil Engineering (CE)
| 1. What is kinematics? |  |
Ans. Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause that motion. It describes the position, velocity, and acceleration of objects as they move.
| 2. How is velocity defined in kinematics? | |
Ans. Velocity is defined as the rate of change of an object's position with respect to time. It is a vector quantity that includes both magnitude (speed) and direction. In kinematics, velocity is often expressed in meters per second (m/s).
| 3. What is the difference between speed and velocity in kinematics? | |
Ans. Speed is a scalar quantity that refers to "how fast an object is moving," whereas velocity is a vector quantity that refers to "the rate at which an object changes its position." While speed only considers the magnitude of motion, velocity takes both magnitude and direction into account.
| 4. How can acceleration be calculated in kinematics? | |
Ans. Acceleration can be calculated by dividing the change in velocity by the change in time. The formula for average acceleration is given by: acceleration = (final velocity - initial velocity) / time. Acceleration is also a vector quantity and is measured in meters per second squared (m/s²).
| 5. What are the three equations of motion in kinematics? | |
Ans. The three equations of motion in kinematics are:
1. v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
2. s = ut + (1/2)at², where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.
3. v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
Text Transcript from Video
okay having understood the time
derivative of the vector let's say our
that we considered earlier let's use
that fact in order to derive the
velocity of a particular rigid body okay
so where do we start from we start from
as usual a fix at frame okay so we
looked at the fixed frame let us say
there is a body at a particular instant
of time okay the question we have
basically is supposing I know the
position of a particular point on the
body I wish to find out what would be
the position of another point any other
arbitrary point that lies on the body
okay so I'm going to just look at the
planar kinematics because that simpler
to understand I mean I can use the board
to draw okay so that's pretty simple
this should be our B okay and what
should I call this s if I have to find
out B in terms of a a becomes a
reference right so it emanates from a
and goes to B are you with me and
therefore if I call this as R it is our
be equal to our a-plus are all these are
vector quantities right re plus R is
equal to RB okay remember we have
defined B with respect to a as you can
see it emanates from a goes to B and
therefore this is the position of B with
respect to a okay an important point to
note okay so I'm going to write the door
here
are of me with respect to so we have
written a relative position of any other
point with respect to one of the
reference points on the solid okay
having done this if I have to find out
the velocity relationship between these
that should be simple all I have to do
is is to take the time derivative of
these okay in that this r dot B equals r
dot a plus or not be with respect to you
okay our B is defined with respect to
the origin of the frame our a is defined
with respect to this whereas our B with
respect to a is a relative position
alright and therefore these two there is
no problem with respect to fix our frame
this will be the velocity of B this will
be the velocity of a and this will be
the velocity of B with respect to you
okay here is where we have to understand
one important thing related to what we
have written here okay and that is this
let me use the same this a screwdriver
supposing I rotate it with a point as
the as the pivotal point right or if I
take this point and rotate it okay let's
see what difference it makes as well as
the rotation itself is concerned okay so
let's say I am going to take a 90 degree
rotation with respect to a central point
it goes like this
all right with respect to an end point
if I take like this it goes like this
all right now supposing I know in this
particular current position one of those
points with respect to a fixed frame and
I know there is a rotation that is
occurring on this rigid body every point
on this rigid body will be seeing the
same rotation if it is seeing the same
rotation it will also see the same rate
of rotation and a double rate of
rotation all right so we're going to use
that fact over here okay so we are
looking at this body rotating okay and
let's say the axis is perpendicular to
this we already derived a relationship
with respect to doing that particular
thing okay so let's make use of that
after having written this now the
question is what is V of B with respect
to Y remember in the earlier exercise we
had a be pivoted about a there was a
rotation occurring all right and we
wanted to find out what would be the
rate of this rotation here the exercise
is similar because given that this point
is fixed and there is a rotation
occurring I would want to find out the
rate of this one important observation
here is how much is it changing so I
have
I have a body that is rotating a rigid
body that is rotating and I have one
point here another point here about
which about a it is rotating what would
be the change in the magnitude bit of RB
with respect to a only the magnitude
zero correct what is the other aspect
that I should look at change in
direction right now when we did that
exercise earlier we looked at change in
direction alone right change in
magnitude the rate of change in
magnitude was considered to be zero
right we separated magnitude and
direction right now we have a situation
where we have to look at only the
directional change of this vector right
so it becomes easy for me if I know that
this is rotating with an angular
velocity Omega then it is nothing but
Omega cross so let me write it in this
time that we have already written if I
have to find out the derivative or dot
of B with respect to a it is omega cross
RB with us 50 days and therefore if I
substitute in this I will get our dot B
equals our dot A plus Omega cross R of B
with
okay so I have written a relation
between two points of the body VB equals
VA plus Omega cross R of B with respect
okay one step and that's it all right
okay let's complicate it a little bit I
want to find out the acceleration okay
what should I do
naturally if I want to find the
acceleration and I'm looking at with
respect to the fixed set for him then
all I have to do is to take derivative
of this particular stuff I simplest that
so let's do that if I take the
derivative so let's say a be
acceleration of B is nothing but V dot B
and that is equal to based on this V dot
a plus Omega cross R B with respect to a
the whole thing the derivative the whole
thing so I'm just putting this at a sort
of line over here on top to represent
that I am taking the entire derivative
okay
now this implies so V dot V equals V dot
a plus what should happen to this I will
have Omega dot cross R B with respect to
a plus Omega cross or dot B with this 50
looks okay
what is omega dot the rate of angular
moment angular velocities angular
acceleration and therefore this will be
acceleration of B can be written as
acceleration of a plus let me just
denote that as alpha angular
acceleration cross or B with respect to
E Plus know you notice that this our dot
B with respect to a is revisiting and I
know that is equal to Omega cross RB I
can just substitute over here so I will
get Omega cross Omega cross RB with
respect to you that's it so I have a B
is given by a at the acceleration at a
plus angular acceleration cross RB with
respect to a plus angular velocity cross
Omega cross RB with respect to Y okay
what is the direction of this and what
is the direction of this if we get an
idea then we are talking a little more
sense all right okay let us just go back
to this okay when I say Omega cross RB
with respect to a Omega is perpendicular
RB with respect to a is like this
perpendicular and this which means if I
draw a line that is perpendicular to
this and axis that is perpendicular to
this this is the direction in which
Omega cross RB with respect to a should
occur so the direction of our dot B with
respect to a is this okay I'm just
showing the direction sorry to reach the
small it is
agree with me this is perpendicular now
observe this expression what's the
direction of Omega cross RB with respect
to e it is the direction of our dot B
with respect to a which is perpendicular
to Omega and RB with respect to a right
now I am taking Omega cross this
particular this is the velocity here
this is the velocity we already know
Omega cross this will give me what
perpendicular to these two but I am
looking at Omega cross this and
therefore it will be in this direction
so this is the direction of Omega cross
Omega cross RB with respect to you you
agree with me I'm just looking at one
component one term in that particular
equation right how about this what's the
derivative of Omega dot the
perpendicular direction is not changing
at all right so if I have theta dot
times n alpha is nothing but theta
double dot times n ok so this is the
direction along the perpendicular to the
board cross RB with respect to a what
should I get so it has to be
perpendicular to those two and that's so
I'll get this and this is alpha cross or
B are you with me now what do I notice
about this particular point if I say see
this direction it is towards a right it
is going to rotate like this so if I
draw an arc about this a a circle about
the C
this seems to be along the radial
direction and this seems to be along the
tangential direction so I'm going to
call this as a what shall I call this as
T of B with respect to e and this I can
call us a n or let's say our AR is fine
a R of B with respect to a so going back
to this I know this is along the
tangential direction so I'm going to
call this a T and this is going to be
along the negative of our direction so I
can call it as a or at a minus re and
therefore acceleration on B is equal to
acceleration P plus a tangential
component and supposing I know that the
body is rotating with a constant angular
velocity immediately alpha dot alpha is
zero okay it is like having a thread so
I am taking this so I'm just rotating
okay with an angular velocity what will
happen there will be a centripetal force
centrifugal force and therefore this
will be holding it in place right now
or in other words this acceleration is
related to basically that it is trying
to ten inward the acceleration is
tending to go inward okay
later on we'll understand how this
translates to a centrifugal force
derivative of the vector let's say our
that we considered earlier let's use
that fact in order to derive the
velocity of a particular rigid body okay
so where do we start from we start from
as usual a fix at frame okay so we
looked at the fixed frame let us say
there is a body at a particular instant
of time okay the question we have
basically is supposing I know the
position of a particular point on the
body I wish to find out what would be
the position of another point any other
arbitrary point that lies on the body
okay so I'm going to just look at the
planar kinematics because that simpler
to understand I mean I can use the board
to draw okay so that's pretty simple
this should be our B okay and what
should I call this s if I have to find
out B in terms of a a becomes a
reference right so it emanates from a
and goes to B are you with me and
therefore if I call this as R it is our
be equal to our a-plus are all these are
vector quantities right re plus R is
equal to RB okay remember we have
defined B with respect to a as you can
see it emanates from a goes to B and
therefore this is the position of B with
respect to a okay an important point to
note okay so I'm going to write the door
here
are of me with respect to so we have
written a relative position of any other
point with respect to one of the
reference points on the solid okay
having done this if I have to find out
the velocity relationship between these
that should be simple all I have to do
is is to take the time derivative of
these okay in that this r dot B equals r
dot a plus or not be with respect to you
okay our B is defined with respect to
the origin of the frame our a is defined
with respect to this whereas our B with
respect to a is a relative position
alright and therefore these two there is
no problem with respect to fix our frame
this will be the velocity of B this will
be the velocity of a and this will be
the velocity of B with respect to you
okay here is where we have to understand
one important thing related to what we
have written here okay and that is this
let me use the same this a screwdriver
supposing I rotate it with a point as
the as the pivotal point right or if I
take this point and rotate it okay let's
see what difference it makes as well as
the rotation itself is concerned okay so
let's say I am going to take a 90 degree
rotation with respect to a central point
it goes like this
all right with respect to an end point
if I take like this it goes like this
all right now supposing I know in this
particular current position one of those
points with respect to a fixed frame and
I know there is a rotation that is
occurring on this rigid body every point
on this rigid body will be seeing the
same rotation if it is seeing the same
rotation it will also see the same rate
of rotation and a double rate of
rotation all right so we're going to use
that fact over here okay so we are
looking at this body rotating okay and
let's say the axis is perpendicular to
this we already derived a relationship
with respect to doing that particular
thing okay so let's make use of that
after having written this now the
question is what is V of B with respect
to Y remember in the earlier exercise we
had a be pivoted about a there was a
rotation occurring all right and we
wanted to find out what would be the
rate of this rotation here the exercise
is similar because given that this point
is fixed and there is a rotation
occurring I would want to find out the
rate of this one important observation
here is how much is it changing so I
have
I have a body that is rotating a rigid
body that is rotating and I have one
point here another point here about
which about a it is rotating what would
be the change in the magnitude bit of RB
with respect to a only the magnitude
zero correct what is the other aspect
that I should look at change in
direction right now when we did that
exercise earlier we looked at change in
direction alone right change in
magnitude the rate of change in
magnitude was considered to be zero
right we separated magnitude and
direction right now we have a situation
where we have to look at only the
directional change of this vector right
so it becomes easy for me if I know that
this is rotating with an angular
velocity Omega then it is nothing but
Omega cross so let me write it in this
time that we have already written if I
have to find out the derivative or dot
of B with respect to a it is omega cross
RB with us 50 days and therefore if I
substitute in this I will get our dot B
equals our dot A plus Omega cross R of B
with
okay so I have written a relation
between two points of the body VB equals
VA plus Omega cross R of B with respect
okay one step and that's it all right
okay let's complicate it a little bit I
want to find out the acceleration okay
what should I do
naturally if I want to find the
acceleration and I'm looking at with
respect to the fixed set for him then
all I have to do is to take derivative
of this particular stuff I simplest that
so let's do that if I take the
derivative so let's say a be
acceleration of B is nothing but V dot B
and that is equal to based on this V dot
a plus Omega cross R B with respect to a
the whole thing the derivative the whole
thing so I'm just putting this at a sort
of line over here on top to represent
that I am taking the entire derivative
okay
now this implies so V dot V equals V dot
a plus what should happen to this I will
have Omega dot cross R B with respect to
a plus Omega cross or dot B with this 50
looks okay
what is omega dot the rate of angular
moment angular velocities angular
acceleration and therefore this will be
acceleration of B can be written as
acceleration of a plus let me just
denote that as alpha angular
acceleration cross or B with respect to
E Plus know you notice that this our dot
B with respect to a is revisiting and I
know that is equal to Omega cross RB I
can just substitute over here so I will
get Omega cross Omega cross RB with
respect to you that's it so I have a B
is given by a at the acceleration at a
plus angular acceleration cross RB with
respect to a plus angular velocity cross
Omega cross RB with respect to Y okay
what is the direction of this and what
is the direction of this if we get an
idea then we are talking a little more
sense all right okay let us just go back
to this okay when I say Omega cross RB
with respect to a Omega is perpendicular
RB with respect to a is like this
perpendicular and this which means if I
draw a line that is perpendicular to
this and axis that is perpendicular to
this this is the direction in which
Omega cross RB with respect to a should
occur so the direction of our dot B with
respect to a is this okay I'm just
showing the direction sorry to reach the
small it is
agree with me this is perpendicular now
observe this expression what's the
direction of Omega cross RB with respect
to e it is the direction of our dot B
with respect to a which is perpendicular
to Omega and RB with respect to a right
now I am taking Omega cross this
particular this is the velocity here
this is the velocity we already know
Omega cross this will give me what
perpendicular to these two but I am
looking at Omega cross this and
therefore it will be in this direction
so this is the direction of Omega cross
Omega cross RB with respect to you you
agree with me I'm just looking at one
component one term in that particular
equation right how about this what's the
derivative of Omega dot the
perpendicular direction is not changing
at all right so if I have theta dot
times n alpha is nothing but theta
double dot times n ok so this is the
direction along the perpendicular to the
board cross RB with respect to a what
should I get so it has to be
perpendicular to those two and that's so
I'll get this and this is alpha cross or
B are you with me now what do I notice
about this particular point if I say see
this direction it is towards a right it
is going to rotate like this so if I
draw an arc about this a a circle about
the C
this seems to be along the radial
direction and this seems to be along the
tangential direction so I'm going to
call this as a what shall I call this as
T of B with respect to e and this I can
call us a n or let's say our AR is fine
a R of B with respect to a so going back
to this I know this is along the
tangential direction so I'm going to
call this a T and this is going to be
along the negative of our direction so I
can call it as a or at a minus re and
therefore acceleration on B is equal to
acceleration P plus a tangential
component and supposing I know that the
body is rotating with a constant angular
velocity immediately alpha dot alpha is
zero okay it is like having a thread so
I am taking this so I'm just rotating
okay with an angular velocity what will
happen there will be a centripetal force
centrifugal force and therefore this
will be holding it in place right now
or in other words this acceleration is
related to basically that it is trying
to ten inward the acceleration is
tending to go inward okay
later on we'll understand how this
translates to a centrifugal force
About this Video
4.68/5
Rating
Oct 04, 2025
Last updated
Related Exams
Kinematics Velocity and Acceleration for Civil Engineering (CE)2025 is part of Civil Engineering (CE) preparation. The notes and questions for Kinematics Velocity and Acceleration
have been prepared according to the Civil Engineering (CE) exam syllabus. Information about Kinematics Velocity and Acceleration covers all important topics
for Civil Engineering (CE)2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises
and tests below for Kinematics Velocity and Acceleration.
Introduction of Kinematics Velocity and Acceleration in English is available as part of our Civil Engineering (CE) preparation & Kinematics Velocity and Acceleration in Hindi
for Civil Engineering (CE) courses. Download more important topics, notes, lectures and mock test series for Civil Engineering (CE) Exam by
signing up for free.
Description
Video Lecture & Questions for Kinematics Velocity and Acceleration Video Lecture - Civil Engineering (CE) - Civil Engineering (CE) full syllabus preparation | Free video for Civil Engineering (CE) exam.
Information about Kinematics Velocity and Acceleration
Here you can find the Kinematics Velocity and Acceleration defined & explained in the simplest way possible. Besides explaining types of Kinematics Velocity and Acceleration
theory, EduRev gives you an ample number of questions to practice Kinematics Velocity and Acceleration tests, examples and also practice Civil Engineering (CE) tests.
Related Searches
video lectures
,Exam
,study material
,Sample Paper
,MCQs
,ppt
,past year papers
,Kinematics Velocity and Acceleration Video Lecture - Civil Engineering (CE)
,Free
,Important questions
,Semester Notes
,Objective type Questions
,Summary
,Extra Questions
,Previous Year Questions with Solutions
,practice quizzes
,Viva Questions
,mock tests for examination
,Kinematics Velocity and Acceleration Video Lecture - Civil Engineering (CE)
,shortcuts and tricks
,Kinematics Velocity and Acceleration Video Lecture - Civil Engineering (CE)
;
Study Kinematics Velocity and Acceleration on the App
Students of Civil Engineering (CE) can study Kinematics Velocity and Acceleration alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Kinematics Velocity and Acceleration,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Kinematics Velocity and Acceleration is prepared as per the latest Civil Engineering (CE) syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup on EduRev and stay on top of your study goals
10M+ students crushing their study goals daily