Question

derive the Kinematic equation of the motion for constant acceleration .​

Answer 1

Answer :-

$\\\;\underbrace{\sf{Understanding\;the\;Question\;:-}}$

Here different relations of Time, Velocity and Acceleration has been used. The different kinematic equations for uniformly accelerated motion are the Three Equations of Motion.

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★ Formula used :-

$\\\;\boxed{\sf{Acceleration,\:a\;=\;\bf{\dfrac{Change\;in\;Velocity}{Time\;Taken}}}}$

$\\\;\boxed{\sf{Average\;Velocity,\:v_{av}\;=\;\bf{\dfrac{Displacement}{Time}}}}$

$\\\;\boxed{\sf{Average\;Velocity,\:v_{av}\;=\;\bf{\dfrac{Total\;Velocity}{2}}}}$

$\\\;\boxed{\sf{Displacement\;=\;\bf{Average\;Velocity\:\times\:Time}}}$

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★ Solution :-

From figure, (Attachment 1)

Let us consider an object moving with constant acceleration along positive direction of X - axis.

• Position of object at instant t = x' = 0
• Position of object at instant t = x
• Velocity of object at instant t = u = 0
• Velocity of object at instant t = v

• Denotions -

» Initial Velocity = u

» Distance = s

» Acceleration = a

» Final Velocity = u

» Time Taken = t

» Change in Velocity = v - u

» Displacement = x - x'

» Total Velocity = v + u

» Distance = Displacement (since body travels in straight line)

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~ For the 1st Equation of Motion :-

By definition, its Velocity after certain time.

Then,

$\\\;\quad\sf{\rightarrow\;\;Acceleration,\:a\;=\;\bf{\dfrac{Change\;in\;Velocity}{Time\;Taken}}}$

$\\\;\quad\sf{\rightarrow\;\;a\;=\;\bf{\dfrac{v\;-\;u}{t\;-\;0}}}$

$\\\;\quad\sf{\rightarrow\;\;a\;=\;\bf{\dfrac{v\;-\;u}{t}}}$

$\\\;\quad\sf{\rightarrow\;\;v\;-\;u\;=\;\bf{at}}$

$\\\;\quad\sf{\rightarrow\;\;v\;=\;\bf{u\;+\;at}}$

$\\\;\qquad\underline{\boxed{\bf{v\;=\;u\;+\;at}}}$

This is the First Equation of Motion popularly known as Velocity - Time relation.

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~ For the 2nd Equation of Motion :-

By definition, distance covered in a certain time.

Average velocity of the object in a time interval 0 to t is given as,

$\\\;\sf{\Rightarrow\;\;Average\;Velocity,\:v_{av}\;=\;\bf{\dfrac{Displacement}{Time}\;=\;\dfrac{x\;-\;x'}{t\;-\;0}}}$

$\\\;\sf{\Rightarrow\;\;x\;-\;x'\;=\;\bf{v_{av}\;\times\;t}}$

Also,

$\\\;\sf{\Rightarrow\;\;Average\;Velocity,\:v_{av}\;=\;\bf{\dfrac{Total\;Velocity}{2}\;=\;\dfrac{v\;+\;u}{2}}}$

Equating these two steps, we get,

$\\\;\sf{\Rightarrow\;\;x\;-\;x'\;=\;\bf{\dfrac{v\;+\;u}{2}\;\times\;t}}$

Now using 1st Equation of Motion,

v = u + at and value of distance (s), we get

$\\\;\quad\sf{\Rightarrow\;\;s\;=\;\bf{\dfrac{2u\;+\;at}{2}\;\times\;t}}$

$\\\;\quad\sf{\Rightarrow\;\;s\;=\;\bf{(\dfrac{2u}{2}\;+\;\dfrac{at}{2})\;\times\;t}}$

$\\\;\quad\sf{\Rightarrow\;\;s\;=\;\bf{ut\;+\;\dfrac{1}{2}at^{2}}}$

$\\\;\qquad\underline{\boxed{\bf{s\;=\;ut\;+\;\dfrac{1}{2}\:at^{2}}}}$

This is the Second Equation of Motion, popularly known as Position - Time relation.

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~ For the 3rd Equation of Motion :-

By definition, Velocity at a certain time.

From first equation of motion : v - u = at

$\\\;\sf{:\mapsto\;\;Displacement\;=\;\bf{Average\;Velocity\:\times\:Time}}$

$\\\;\sf{:\mapsto\;\;x\;-\;x'\;=\;\bf{\dfrac{v\;+\;u}{2}\:\times\:t}}$

$\\\;\sf{:\mapsto\;\;v\;-\;u\;=\;\bf{\dfrac{2}{t}\:(x\;-\;x')}}$

Multiplying First Equation of Motion, and Distance(s) we get,

$\\\;\quad\sf{:\mapsto\;\;(v\;-\;u)\:(v\;+\;u)\;=\;\bf{\dfrac{2}{t}\:(s)\:at}}$

$\\\;\quad\sf{:\mapsto\;\;v^{2}\;-\;u^{2}\;=\;\bf{2a(s)}}$

$\\\;\qquad\underline{\boxed{\bf{v^{2}\;=\;u^{2}\;+\;2as}}}$

This is the Third Equation of Motion, popularly known as Velocity - Position relation.

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★ More to know :-

We generally heard of Three Equations of Motion. But there is one more equation of motion derived from Second Equation of Motion.

Let's discuss its derivation also.

Refer Attachment 2, for figure of this.

✒ Distance travelled in first t sec is given by

$\\\;\tt{s_{t}\;=\;ut\;\;+\;\;\dfrac{1}{2}\:at^{2}}$

✒ Distance travelled in first (n - 1) sec is

$\\\;\tt{s_{n\;-\;1}\;=\;u(n\;-\;1)\;\;+\;\;\dfrac{1}{2}\:a(n\;-\;1)^{2}}$

✒ Distance travelled in first n sec is

$\\\;\tt{s_{n}\;=\;un\;\;+\;\;\dfrac{1}{2}\:an^{2}}$

Hence, the distance travelled in n th second is,

$\\\;\tt{\leadsto\;\;s_{n_{th}}\;=\;s_{n}\;-\;s_{n\;-\;1}}$

$\\\;\tt{\leadsto\;\;s_{n_{th}}\;=\;(un\;+\;\dfrac{1}{2}an^{2})\;-\;[u(n\;-\;1)\;+\;\dfrac{1}{2}a\:(n\;-\;1)^{2}]}$

$\\\;\tt{\leadsto\;\;s_{n_{th}}\;=\;(un\;+\;\dfrac{1}{2}an^{2})\;-\;[un\;-\;u\;+\;\dfrac{1}{2}a\:(n^{2}\;-\;2n\;+\;1)]}$

$\\\;\tt{\leadsto\;\;s_{n_{th}}\;=\;un\;+\;\dfrac{1}{2}an^{2}\;-\;un\;+\;u\;-\;\dfrac{1}{2}an^{2}\;+\;an\;-\;\dfrac{a}{2}}$

$\\\;\qquad\boxed{\bf{s_{n_{th}}\;=\;u\;+\;\dfrac{a}{2}(2n\;-\;1)}}$