Analytical and graphical method of proof equations of motion
Answers
Suppose a body is moving with an initial velocity (u) with an uniform acceleration (a) for time (t).
Now, we know,
Acceleration= rate of change in velocity/ time
Or, acceleration= initial velocity - final velocity/ time
or, a = v - u / t
or, v - u = at
or, v = u + at [1st eq. of motion]
Analytical method to derive 2nd eq. of motion:
Suppose, a body is traveling with a initial velocity (u) with an uniform acceleration (a) for time (t).
Here, to derive 2nd eq. of motion, we require the formula of average velocity.
We know, average velocity =
initial velocity + final velocity / 2 × t
Therefore, av. velocity = u + v / 2 × t ....... (i)
We know, from the 1st eq. of motion ,
v = u + at
Therefore, putting the value of v in eq. (i)
s = u + u + at / 2 × t (s = distance)
s = 2u + at /2 × t
s = (2u/2 + at/2 ) × t
s = (u + 1/2 at) × t
s = ut + 1/2at sq. [2nd eq. of motion]
Deriving 3rd eq. of motion
We know, from the 1st eq. of motion ,
v = u + at
Therefore, squaring both sides,
v sq. = ( u + at) sq.
v sq. = u sq. + 2uat + a sq. t sq.
v sq. = u sq. + 2a ( ut + 1/2at sq.)
v sq. = u sq. + 2as [3rd eq. of motion]
Note : Please note that 2 (square) should be written on top of the digits not sq. It was for my convenience.
Answer:
Explanation:Analytical method to derive 1st eq. of motion:
Suppose a body is moving with an initial velocity (u) with an uniform acceleration (a) for time (t).
Now, we know,
Acceleration= rate of change in velocity/ time
Or, acceleration= initial velocity - final velocity/ time
or, a = v - u / t
or, v - u = at
or, v = u + at [1st eq. of motion]
Analytical method to derive 2nd eq. of motion:
Suppose, a body is traveling with a initial velocity (u) with an uniform acceleration (a) for time (t).
Here, to derive 2nd eq. of motion, we require the formula of average velocity.
We know, average velocity =
initial velocity + final velocity / 2 × t
Therefore, av. velocity = u + v / 2 × t ....... (i)
We know, from the 1st eq. of motion ,
v = u + at
Therefore, putting the value of v in eq. (i)
s = u + u + at / 2 × t (s = distance)
s = 2u + at /2 × t
s = (2u/2 + at/2 ) × t
s = (u + 1/2 at) × t
s = ut + 1/2at sq. [2nd eq. of motion]
Deriving 3rd eq. of motion
We know, from the 1st eq. of motion ,
v = u + at
Therefore, squaring both sides,
v sq. = ( u + at) sq.
v sq. = u sq. + 2uat + a sq. t sq.
v sq. = u sq. + 2a ( ut + 1/2at sq.)
v sq. = u sq. + 2as [3rd eq. of motion]
Note : Please note that 2 (square) should be written on top of the digits not sq. It was for my convenience.