9. Physics | MOD | Cases of Uniform Acceleration | by Ashish Arora (GA)

# 9. Physics | MOD | Cases of Uniform Acceleration | by Ashish Arora (GA)

now we are, going to discuss the, motion analysis
with uniform acceleration . say when a particle moves with uniform acceleration, in rectilinear
motion we can simply state, say. When a particle. Moves at. Constant acceleration (a). and it
starts with, initial velocity, u. see we are required to find velocity at a general time,
you can simply state, the velocity of particle, after time t can be. Obtained as. We simply
write acceleration as d-v by d-t. and here we can write d-v=(a) d-t. now we integrate
this expression, at t=zero we can substitute the limit of velocity was u, in the beginning
it starts with initial velocity u which was given, at a general time t the velocity of
particle see becomes v. and on integration lefthand-side will become v-minus-u and on
righthand-side we can say, as acceleration is constant it can be taken out, and it becomes
integration from zero to t that will becomes just t, velocity after time t can be written
as, u+(a-t). so you must be very carefully about this equation equation number one, we
call. first speed equation. For. Constantly accelerated motion. or constant accelerated
motion we can directly use this expression which gives as velocity after time t if initial
velocity is, u. further if we are going to use the same expression,
like see we write, further we use, if we are given as v=u+ (a-t), we can simply write
v as d-x by d-t. which is u+ (a-t) and we can write d-x=u + (a-t) into d-t. this gives
us the displacement travel in a duration d-t if we want to calculate the displacement after
time t, so we can say, initially when time was zero the displacement was zero, when the
particle was just started. and at a general time t particle will be at position say s.
and on integration we get integration of d-x, that will be s. and if you integrate u+ (a-t)
then u and (a) are constants so it can be written as u-t +, (a-t)-square by, two, so
this s=u-t+, half a-t-square. This equation which is equation number two, this gives us
the displacement covered by the particle in a time t , if it is moving at a constant acceleration
(a) and with an initial velocity u, this equation we termed as. second speed equation. For.
Uniformly accelerated motion. Let’s discuss another equation that is speed equation number
three, or third speed equation for uniformly accelerated motion. Similarly. If we use. Acceleration as v d-v
by d-x. here we can write v d-v=(a) d-x. on integrating this expression, say if at
starting point the displacement was zero and velocity was u, say after covering displacement
s, say velocity becomes v. and on integrating we get the final velocity after traveling
the displacement s, so here it will be v-square by two, and we apply limits from u to v and
on righthand-side we get (a) into s. so here lefthand-side will become. V-square minus
u-square by two=(a-s), or we can write v-square=u-square+ two (a-s). here this equation gives
us the velocity of particle . after covering a displacement, s. if it was moving with an
uniform acceleration (a) and with initial speed was started with u. this equation we
write. Third speed equation, for, uniformly accelerated motion. Let’s discuss the displacement of a particle
, during n-eth second for uniformly accelerated motion. Say for example a particle starts
from time t=zero, and it is moving in a rectilinear motion, see if we talk the time from t=zero
to t=one second. This we can call as. First second of motion, from time t=one to t=two
again the duration is one second this we call, second second-of motion similarly from t=
two to t=three it is the, third second-of motion similarly we can define different motions.
If a particle starts, with an initial speed u, and it is accelerating with acceleration
a, so we can say after every instant it’s velocity is gradually increasing. As it is
gradually increasing. we can state, that later on after starting, in one second duration
particle will cover more distance or displacement compare to the, initial duration of one second.
Say we find out the displacement. covered. In n-eth second. Of motion. So n-eth second
of motion if say particle covered a displacement delta-s-n, it can be written displacement
covered up to n second. Minus displacement covered up to n minus one-eth second. So we
just subtract the displacement up to n minus one-eth second from, the displacement which
is covered up to n-eth second, so it will give us the displacement, delta-s which is
covered, during n-eth second, and if we talk about n-eth second. This is the time from
t=n minus one to t=n, so. For calculation of this displacement covered in n-eth second,
we are finding the displacement in n second and n minus one second and we are taking the
difference, so this can be easily calculated by speed equation number two i.e. u-t + half-a-t-square.
For n-eth second we can write it , u-n, plus, half-a-n-square. Minus, we subtract it the
displacement up to n minus one-eth second i.e. u into n-minus-one+, half-(a) into n-minus-one
whole-square. When we open this term, we get u-n +, half-a-n-square, this minus u-n, +
u, than we are getting minus half (a)-n-square, than minus, half-(a)+ (a)-n so here if you
are just carefully about look on it this half (a-n)- square gets cancelled out, this u-n
gets cancelled out, what will be left is u. + we can take half-a common, this will be
half-a into two-n minus, one. This is the relation used for calculation of . displacement
covered by the particle during, n-eth second, this is also quite an important relation,
so just keep it in mind in various problem it is going to be use directly. Now let’s write on a note, which will be,
extremely helpful for us, to analyze speed equations in various problems. In this note
, we are going to write. All the speed equations. Are. Vector equations. And. The motion parameters.
Like displacement. Velocity final velocity, initial velocity. Acceleration. Are taken.
With appropriate. Signs. According to the. Reference frame chosen. So always be careful.
While applying this speed equations in various cases these all are to be used as vector equations,
as using these equations as vector equations , will always be a beneficial way to handle
a problem and, by this method using this vector equation we can also reduce the length of
. solution of a problem. Like if we talk about these equation the first equation we use as
v=u+, (a-)t. the second equation is s is, u-t +, half(a)-t-square, third is v-square=
u-square + two,(a) dot s, and forth which we can use we have just now, derive i.e.,
displacement in n-eth second i.e. , u+ (a) by two, two-n minus one so be careful about
these equations in various cases we are going to use these equations as vector equations.

## One Reply to “9. Physics | MOD | Cases of Uniform Acceleration | by Ashish Arora (GA)”

1. Aditi Mukherjee says:

sir equation no. 3 is a dot product equation, therefore its a scalar equation