Question

ahoy!!
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derive equation for displacement in nth second of motion.

explanation required.
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thank you.

Answer 1

good afternoon.
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time: 2:19
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suppose there is an object which starts moving with initial velocity u and is having the uniform acceleration a.
let the displacement of the object in the nth second of its motion be $s_{nth}$

then, 

[tex] s_{nth} = s_{n} - s_{n-1} .............................(1) [/tex]

where $s_{n}$ and $s_{n-1}$ are displacements of the object in n and n-1 seconds.

the values of $s_{n}$ and $s_{n-1}$ can be obtained by putting t=n and t= n-1 in the equation (1)

we know,

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

therefore the equation becomes:

$s_{nth} = [ un + \frac{1}{2} a n^{2} ] - [ u (n-1)+ \frac{1}{2} a (n-1)^{2} ]$


[tex] s_{nth} = [un + \frac{1}{2} a n^{2} ]-[u(n-1)+ \frac{1}{2} a ( n^{2} +1-2n)] [/tex]

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

$un -un + u + \frac{1}{2}an^2 - \frac{1}{2} an^2 - \frac{1}{2} a + \frac{1}{2} 2an$

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

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

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end.

Answer 2

$\textbf{Answer is in Attachment !}$