Question

Prove that Third equation of motion Also gives the name of each parameter used in it.​

Answer 1

Explanation:

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

Derivation of Third Equation of Motion by Calculus Method

We know that acceleration is the rate of change of velocity and can be represented as:

a=dvdt (1)

We also know that velocity is the rate of change of displacement and can be represented as:

v=dsdt (2)

Cross multiplying (1) and (2), we get

adsdt=vdvdt

∫s0ads=∫vuvds

as=v2−u22

v2=u2+2as

This is how we derive the three equations of motion by algebraic, graphical and calculus method.

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