Question

give the derivation of second and third equation of motion​

Answer 1

Pls note- This pic is for derivation of 2nd and 3rd equation of motion.

Answer:

First, consider a body moving in a straight line with uniform acceleration. Then, let the initial velocity be u, acceleration be a, time period be t, velocity be v, and the distance travelled be S.

Derivation of First Equation of Motion by Algebraic Method-

It is known that the acceleration (a) of the body is defined as the rate of change of velocity.

So, the acceleration can be written as:

a = v − ut

From this, rearranging the terms, the first equation of motion is obtained, which is:

v = u + at______equation1.

Derivation of 2and equation of morion-

Refer to diagram attached in the beginning.

In this diagram, the distance travelled (S) = Area of figure OABC = Area of rectangle OADC + Area of triangle ABD.

Now, the area of the rectangle OADC = OA × OC = ut

And, Area of triangle ABD = (1/2) × Area of rectangle AEBD = (1/2) at2 (Since, AD = t and BD = at)

Thus, the total distance covered will be:

S = ut + (1/2) at2_______equation2.

Derivation of Third Equation of Motion by Algebraic Method-

For this, consider the same diagram as for 2nd equation.

The total distance travelled, S = Area of trapezium OABC.

So, S= 1/2(SumofParallelSides)×Height

S=(OA+CB)×OC

Since, OA = u, CB = v, and OC = t

The above equation becomes

S= 1/2(u+v)×t

Now, since t = (v – u)/ a

The above equation can be written as:

S= 1/2(u+v)×(v-u)/a

Rearranging the equation, we get

S= 1/2(v+u)×(v-u)/a

S = (v2-u2)/2a

Third equation of motion is obtained by solving the above equation:

v2 = u2+2aS______equation3.

Hope the answer helps u

Thank u