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Answer 1

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Derivation of Second Equation of Motion by Graphical Method

Derivation of Equation of Motion

From the graph above, we can say that

Distance travelled (s) = Area of figure OABC = Area of rectangle OADC + Area of triangle ABD

s=(12AB×BD)+(OA×OC)

Since BD = EA, the above equation becomes

s=(12AB×EA)+(u×t)

As EA = at, the equation becomes

s=12×at×t+ut

On further simplification, the equation becomes

s=ut+12at2

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

ds=(u+at)dt=(udt+atdt)

On further simplification, the equation becomes:

∫s0ds=∫t0udt+∫t0atdt

=ut+12at2

Derivation of Third Equation of Motion

Derivation of Third Equation of Motion by Algebraic Method

We know that, displacement is the rate of change of position of an object. Mathematically, this can be represented as:

Displacement=(InitialVelocity+FinalVelocity2)×t

Substituting the standard notations, the above equation becomes

s=(u+v2)×t

From the first equation of motion, we know that

v=u+at

Rearranging the above formula, we get

t=v−ua

Substituting the value of t in the displacement formula, we get

s=(v+u2)(v−ua)

s=(v2−u22a)

2as=v2−u2

Rearranging, we get

v2=u2+2as

Derivation of Third Equation of Motion by Graphical Method

From the graph, we can say that

The total distance travelled, s is given by the Area of trapezium OABC.

Hence,

S = ½ (Sum of Parallel Sides) × Height

S=(OA+CB)×OC

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

The above equation becomes

S= ½ (u+v) × t

Now, since t = (v – u)/ a

The above equation can be written as:

S= ½ ((u+v) × (v-u))/a

Rearranging the equation, we get

S= ½ (v+u) × (v-u)/a

S = (v2-u2)/2a

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

v2 = u2+2aS

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Answer 2

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