Question

A particle starts its motion with velocity u and moves with uniform acceleration a for n second then find out distance travelled in n th second?​

Answer 1

Answer:

$u + \frac{a}{2} (2n - 1)$

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Given

Initial Velocity = 'u'

Distance travelled = 'S'

Uniform Acceleration = 'a'

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We Know that

$S = ut + \frac{1}{2} a {t}^{2}$

In the same way in 'n' seconds

$Sn = un + \frac{1}{2} a {n}^{2}$

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According To Question

$S(n - 1) = u(n - 1) + \frac{1}{2} a {(n - 1)}^{2}$

&

$Snth = Sn - S(n - 1)$

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Now on putting all values

$= > Snth = un + \frac{a {n}^{2} }{2} - [un - u + \frac{a}{2} {(n - 1)}^{2}] \\ = > Snth = \cancel{un} + \frac{a {n}^{2} }{2} - \cancel{un} + u - \frac{a}{2} ( {n}^{2} + 1 - 2n)$

On Solving brackets

$= > Snth = \cancel{\frac{u {n}^{2} }{2} } + u - \cancel{\frac{u {n}^{2} }{2}} - \frac{a}{2} + \frac{2na}{2} \\ = > Snth = u - \frac{a}{2} + \frac{2na}{2}$

On Taking a/2 as Common

$\boxed{Snth = u + \frac{a}{2} (2n - 1)}$

Where

'n' is a natural number.

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Used Identity In the Solution

${(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab$

Answer 2

Answer:

particle velocity=u moves with acceleration a

=>distance travelled in n second =

$un + \frac{a {n}^{2} }{2}$

=>distance travelled in n-1 second =

$un - u + \frac{(a )({n - 1})^{2} }2{} \\ \\ un - u + \: \frac{a {n}^{2} }{2} + \frac{a}{2} - an$

distance travelled in n th second = u-a/2 + an