Question

Derive second equation of motion by calculus method?

Answer 1

Answer:

$\bullet \; \large{\tt S = ut + \dfrac{1}{2} \; at^2}$

To Prove:

• Second Kinematic Equation.

Explanation:

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Velocity:

It is defined as rate of change of Displacement w.r.t time.

S.I units: m/s

Expression:

$\qquad \large{\tt v = \dfrac{\Delta s}{\Delta t}}$

Talking over a small interval of time,

$\longmapsto \large{\tt v = \dfrac{ds}{dt}}$

Rearranging,

$\longmapsto \large{\tt ds = v.dt}$

Taking over a large time.

$\longmapsto \large{\tt \displaystyle\int \tt ds = \displaystyle\int \tt v.dt}$

Applying Limits,

$\longmapsto \large{\tt \displaystyle\int\limits_{0}^{s} \tt ds = \displaystyle\int\limits_{0}^{t} \tt v.dt}$

But from first Kinematic Equation we Know, [v = u + at]

Substituting,

$\longmapsto \large{\tt \displaystyle\int\limits_{0}^{s} \tt ds = \displaystyle\int\limits_{0}^{t} \tt (u + at)dt}$

$\longmapsto \large{\tt \displaystyle\int\limits_{0}^{s} \tt ds = \displaystyle\int\limits_{0}^{t} \tt u.dt + \displaystyle\int\limits_{0}^{t} \tt at.dt}$

Simplifying,

$\longmapsto \large{\tt \bigg[s \bigg]^s_0 = \bigg[ut \bigg]^t_0 + \bigg[\dfrac{at^2}{2} \bigg]^t_0 }$

$\longmapsto \large{\tt S - 0 = ut + \dfrac{1}{2} \; at^2}$

$\longmapsto \large{\underline{\boxed{\red{\tt S = ut + \dfrac{1}{2} \; at^2}}}}$

Hence Derived !!

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Special Cases:

1. If Distance is zero i.e. ( S = 0) then equation becomes, u t = - 1/2 a t².
2. If initial velocity is zero i.e. ( u = 0) then equation becomes, S = 1/2 a t².
3. If Acceleration is zero i.e. ( a = 0) then equation becomes,S = u t.
4. If time is zero i.e. ( t = 0) then equation becomes, S = 0.

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Answer 2

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