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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
$\frac{{2\upsilon \left( {n + 1} \right)}}{n}$
$\frac{{\upsilon \left( {n + 1} \right)}}{n}$
$\frac{{\upsilon \left( {n - 1} \right)}}{n}$
$\frac{{2\upsilon \left( {n - 1} \right)}}{n}$
Solution
(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}$
$\therefore$ Distance travelled in last two seconds,
$ = {S_n} – {S_{n – 2}}$$ = \frac{1}{2}a{n^2} – \frac{1}{2}a{(n – 2)^2}$
$ = \frac{a}{2}\left[ {{n^2} – {{(n – 2)}^2}} \right]$$ = \frac{a}{2}[n + (n – 2)][n – (n – 2)]$
=$a(2n – 2)$ $ = \frac{v}{n}(2n – 2)$$ = \frac{{2v(n – 1)}}{n}$
