Dimension of S=vit+1/2gt²
Answers
Answer:
Trying to simplify this to make it more familiar for everyone.
Motion Under Gravity:
Scenario: Upward Motion
v = u - at
h = ut - 1 / 2 gt^2
v^2 - u^2 - 2gh
Scenario: Downward Motion
v = u + at
h = ut + 1 / 2 gt^2
v^2 - u^2 + 2gh
And in Gen:
v = u + at
h = ut + 1 / 2 at^2
v^2 - u^2 + 2as
Now lets assume Distance traveled by a body in nth second:
sn = u + (2n - 1) a / 2
Now, s = distance traveled
h = height
t = time
u = initial velocity
v = final velocity
a = acceleration
g = gravitational acceleration
If a body starts from rest and falls freely or is dropped, then u = 0.
If the body is thrown upwards then it will rise until its vertical velocity becomes zero. Maximum height attained is h = u2 /2g.
If a packet is dropped at a height h, from an aero-plane or balloon ascending with a velocity u, then the time taken by the packet to reach ground is given by h = ut + 1/2 gt^2
No line in x - t graph can be perpendicular to time axis because it will represent infinite velocity.
If the x - t graph is a curve whose slope decreases continuously with time, then the velocity of the body goes on decreasing continuously and the motion of the body is retarded.
If the x - t graph is a curve whose slope continuously increases, then the velocity of the body is continuously increasing and the body accelerated.
If the v - t graph is a straight line parallel to time axis, then the acceleration of the body is zero.
If the u - t graph is a straight line inclined to time axis with positive slope, then that body is moving with constant acceleration.
If v - t graph is a straight line inclined to time axis with negative slope, then the body is retarded.
The v - t graph normal to time axis is not a practical possibility because it means that the acceleration of the body is infinite.
For a body moving with uniform acceleration, the average velocity = (u + v)/2, where u is the initial velocity
Explanation:
so it's helpful for you