Question

Derive all three equation of motion using general method and graphical method.

Answer 1

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Draw a line parallel to x-axis DA from point, D from where object starts moving. Draw another line BA from point B parallel to y-axis which meets at E at y-axis. Above equation is the relation among initial vlocity (u ), final velocity (v ), acceleration (a) and time (t). It is called first equation of motion.

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Answer 2

Answer:

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).

Following details are obtained from the graph above:

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:

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 Second Equation of Motion by Calculus Method

Velocity is the rate of change of displacement.

Mathematically, this is expressed as

v=dsdt

Rearranging the equation, we get

ds=vdt

Substituting the first equation of motion in the above equation, we get

ds=(u+at)dt =(udt+atdt) ∫s0ds=∫t0udt+∫t0atdt s=ut+12at2

Derivation of Third Equation of Motion

The third equation of motion is:

v2 = u2 + 2aS

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

Also I have inserted picture..

The first picture is of the first method, the second picture is of the second method and third picture is for the third equation..

These all are graphical method... If u want the general method then again u have to ask the question because in this I can't able to fit both the method..

Hope it helps!! Pls mark as brainlist and follow me .