Question

${ \large{ \boxed{ \red{ \underline{ \bf \: Derive \: the \: 3rd \: equation \: of motion}}}}}$
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Answer 1

Answer:

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Explanation:

$\star$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=(Initial Velocity + Final Velocity/2)×t

Substituting the standard notations, the above equation becomes

s=(u+v/2)×t

From the first equation of motion, we know that

v=u+at

Rearranging the above formula, we get

t=v−u/a

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

s=(v+u/2)(v−u/a)

s=(v² −u²/2a)

2as=v²−u²

Rearranging, we get

v²=u²+2as

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$\star$Derivation of Third Equation of Motion by Graphical Method

see image

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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$\star$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=dv/dt ---- (1)

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

v=ds/dt ---- (2)

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

$a \frac{ds}{dt} = v\frac{dv}{dt} \\ \\ {∫}^{s}_0ads = {∫}^{u}_vvds \\ \\ as = \frac{ {v}^{2} - {u}^{2} }{2}$

v²=u²+2as

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