Hello friends. Today I am going to explain Equations of straight line motion at constant
acceleration and we will see how these equations can be used in real life cases. In order to
derive the equations of straight line motion at constant acceleration consider an example
where a block has an initial velocity of u and after t time lets say it has reached a
velocity of v, assuming a constant acceleration is maintained by the person and let S be the
displacement of the block. We have already learned about acceleration, velocity and displacement
in the previous videos, if you are not clear about them, please check them out (see description).
So the acceleration a= dv/dt or delta-v by delta t which is final velocity -initial velocity
by the time taken. On simplifying this equation we will get v=u+at. With this equation, we
can find final velocity from given initial velocity, acceleration and time. Since velocity
is given by v=dx/dt, if we substitute this in equation 1 and on integrating this equation
with proper limits we can find displacement in terms of initial velocity, acceleration
and time which is given by s=ut+1/2at^2 and there is one more equation which we can get
by squaring equation 1 and on simplifying this and using equation 2 we can get the relation
v^2-u^2=2as. So these 3 equations are called as equations of motion for a straight line
motion at constant acceleration. Now let us apply these equations on a freely falling
problem.Let us say a pokemon ball is dropped from the top of the building and let say it
hits the ground after t seconds, the velocity of the ball increases in the downward direction
at a constant rate because the acceleration due to the gravity is a constant value.If
the ball is just left from the top of the building without giving any additional velocity
then we can consider intial velocity to be zero and if we take the y axis in the upward
direction then acceleration will be negative g and the value of acceleration due to gravity
for earth is 9.8 m/s^2 and just before impact with the ground, the ball has velocity v in
the negative y direction so we can find that v=gt and the time taken to reach the ground
is given by v/g which is also called as time of descent. Now take the second equation,
here displacement is equal to the height of the building and its direction is along the
negative Y axis. So s=-H and we can find the value of H as 1/2gt^2 from which we can find
the time of descent as Sqrt(2h/g) and from these two equations we can find that V = Sqrt(2gh).
So we can say that as the height of the building increases the velocity of impact will increase
and the time of descent will also increase with the height of the building.Please note
that theses equations are derived without considering air resistance. Now let us consider
another case where the ball is thrown straight up with a velocity of u m/s, in this case,
the ball will reach a maximum height H and after reaching the maximum height it will
return back as a freely falling body at maximum height velocity is equal to zero and lets
say the velocity when it returns back is V1 and the time taken to goto maximum height
is called time of ascent (ta) and the time taken to return back from maximum height is
called time of descent. So writing the equations of motion during the time of ascent i.e. with
intial velocity u and final velocity zero, we can get time of ascent = u/g and the maximum
height reached by the ball can be given by this equation or we can use another equation
v^2-u^2=2as and we can find H as u^2/2g. Both of these equations will give the same value
of H. So the return motion of the body will be same as that of freely falling body case,
since it is already solved we can use those equations. So eliminating maximum height H
from these 2 equations we can get that u=v, which means the velocity you have given to
the ball intially we be equal to the return velocity of the ball. For example, if you
have thrown a ball at 4 m/s then it comes back with the same velocity 4 m/s but the
direction will be opposite and also we can observe that time of ascent = time of descent.
So please take a look at the summary and try to solve the practice problems. Thank You
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