A body starts from rest with uniform accelera | vedclass

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(d) Now, distance travelled in $n$ sec. ==> ${S_n} = \frac{1}{2}a{n^2}$and distance travelled in $(n - 2)\sec $==>${S_{n - 2}} = \frac{1}{2}a{(n

Vedclass · Accepted Answer

$\frac{{2\upsilon \left( {n - 1} \right)}}{n}$

Vedclass · Answer

(d) Now, distance travelled in $n$ sec. ==> ${S_n} = \frac{1}{2}a{n^2}$and distance travelled in $(n - 2)\sec $==>${S_{n - 2}} = \frac{1}{2}a{(n - 2)^2}$

$	herefore$ Distance travelled in last two seconds,

$ = {S_n} - {S_{n - 2}}$$ = \frac{1}{2}a{n^2} - \frac{1}{2}a{(n - 2)^2}$

$ = \frac{a}{2}\left[ {{n^2} - {{(n - 2)}^2}} ight]$$ = \frac{a}{2}[n + (n - 2)][n - (n - 2)]$

=$a(2n - 2)$ $ = \frac{v}{n}(2n - 2)$$ = \frac{{2v(n - 1)}}{n}$

A body starts from rest with uniform acceleration. If its velocity after $n$ second is $\upsilon ,$ then its displacement in the last two seconds is

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