Question

derive third equation of motion​

Answer 1

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Explanation:

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Answer 2

Answer:

The third equation of motion : v² - u² = 2as.

Derivation of third equation of motion :

• Algebraic method -

We know, a = (v - u)/t

⇒ t = (v - u)/a

Using second equation of motion :

⇒ s = ut + 1/2 at²

⇒ s = u [(v - u)/a] + 1/2 a[(v - u)/a]²

⇒ s = (uv - u²)/a + a(v² + u² - 2uv)/2a²

⇒ s = (uv - u²)/a + (v² + u² - 2uv)/2a

⇒ s = (2uv - 2u² + v² + u² - 2uv)/2a

⇒ s = (v² - u²)/2a

⇒ 2as = v² - u²

• Calculus Method -

Acceleration = dv/dt

Multiplying ds in numerator and denominator,

[tex]\longrightarrow \tt{a=\dfrac{dv}{dt}\times\dfrac{ds}{ds}} [/tex]

[tex]\longrightarrow \tt{a=\dfrac{dv}{ds}\times\dfrac{ds}{dt}} [/tex]

[tex]\longrightarrow \tt{a=v\dfrac{dv}{ds}}} [/tex]

[tex]\longrightarrow \tt{a\,ds = v\,dv}} [/tex]

[tex]\longrightarrow \tt{\int\limits^s_0 {a} \, ds = \int\limits^v_u {v} \, dv} [/tex]

[tex]\longrightarrow \tt{\int\limits^s_0 {a} \, ds = \bigg[\dfrac{v^2}{2}\bigg]^v_u} [/tex]

[tex]\sf\bigg[By\,using : \int\limits {x}^n \, dx = \dfrac{{x}^{n+1}}{n+1}\bigg]} [/tex]

[tex]\longrightarrow\tt{a[s]^s_0=\dfrac{v^2}{2}-\dfrac{u^2}{2}} [/tex]

[tex]\longrightarrow\tt{a(s-0)=\dfrac{v^2-u^2}{2}} [/tex]

[tex]\longrightarrow\tt{2a(s-0)=v^2-u^2} [/tex]

Hence, 2as = v² - u².