Question

Do derivation of equation of motion- (of 1st, 2nd and 3rd) by
graphical method.​

Answer 1

Explanation:

Derivation of First Equation of Motion by Graphical Method

Consider the diagram of the velocity-time graph of a body below:

Derivation Of Equation Of Motion

In this, the body is moving with an initial velocity of u at point A. The velocity of the body then changes from A to B in time t at a uniform rate. In the above diagram, BC is the final velocity i.e. v after the body travels from A to B at a uniform acceleration of a. In the graph, OC is the time t. Then, a perpendicular is drawn from B to OC, a parallel line is drawn from A to D, and another perpendicular is drawn from B to OE (represented by dotted lines).

The initial velocity of the body, u = OA

The final velocity of the body, v = BC

From the graph,BC = BD + DC

So, v = BD + DC

v = BD + OA (since DC = OA)

Finally, v = BD + u (since OA = u) (Equation 1)

Now, since the slope of a velocity-time graph is equal to acceleration a,

So,

a = slope of line AB

a = BD/AD

Since AD = AC = t, the above equation becomes:

BD = at (Equation 2)

Now, combining Equation 1 & 2, the following is obtained:

v = at + u

Derivation of Second Equation of Motion by Graphical Method

Taking the same diagram used in first law derivation:

Derivation Of Equation Of Motion

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

Derivation of Third Equation of Motion by Graphical Method

Derivation Of Equation Of Motion

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

i hope it will be helpful

Answer 2

Answer:

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