Differentiate-s-ut-1-2-at-2-from-the-first-principle-

Question Number 6152 by sanusihammed last updated on 16/Jun/16

D i f f e r e n t i a t e s = u t + \frac{1}{2} {a t}^{2} f r o m t h e f i r s t p r i n c i p l e

Answered by Rasheed Soomro last updated on 16/Jun/16

s = u t + \frac{1}{2} {a t}^{2}

s + δ s = u (t + δ t) + \frac{1}{2} a {(t + δ t)}^{2}

δ s = (u (t + δ t) + \frac{1}{2} a {(t + δ t)}^{2}) - (u t + \frac{1}{2} {a t}^{2})

= u (t + δ t) + \frac{1}{2} a {(t + δ t)}^{2} - u t - \frac{1}{2} {a t}^{2}

= u (t + δ t - t) + \frac{1}{2} a (t^{2} + 2 t δ t + δ t^{2} - t^{2})

= u δ t + \frac{1}{2} a (2 t δ t + δ t^{2})

\frac{δ s}{δ t} = u + a t + \frac{1}{2} a δ t

\underset{δ t \to 0}{l i m} \frac{δ s}{δ t} = \underset{δ t \to 0}{l i m} (u + a t + \frac{1}{2} a δ t)

\frac{d s}{d t} = u + a t

Answered by FilupSmith last updated on 16/Jun/16

f (t) = u t + \frac{1}{2} {a t}^{2}

f (t + h) = u (t + h) + \frac{1}{2} a {(t + h)}^{2}

f^{'} (t) = \lim_{h \to 0} \frac{f (t + h) - f (t)}{h}

f^{'} (t) = \lim_{h \to 0} \frac{(u (t + h) + \frac{1}{2} a {(t + h)}^{2}) - (u t + \frac{1}{2} {a t}^{2})}{h}

f^{'} (t) = \lim_{h \to 0} \frac{u (t + h - t) + \frac{1}{2} a ({(t + h)}^{2} - t^{2})}{h}

f^{'} (t) = \lim_{h \to 0} \frac{u (h) + \frac{1}{2} a (t^{2} + 2 t h + h^{2} - t^{2})}{h}

f^{'} (t) = \lim_{h \to 0} \frac{u h + \frac{1}{2} a (2 t h + h^{2})}{h}

f^{'} (t) = \lim_{h \to 0} \frac{u h + \frac{1}{2} a h (2 t + h)}{h}

f^{'} (t) = \lim_{h \to 0} (u + \frac{1}{2} a (2 t + h))

f^{'} (t) = u + \frac{1}{2} a (2 t)

∴ f^{'} (t) = u + a t

Commented by FilupSmith last updated on 16/Jun/16

added this as a different notation

Commented by sanusihammed last updated on 16/Jun/16

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