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(A) Given that the body starts from rest,initial velocity $u = 0$. Using the first equation of motion,$v = u + at$,we get $v = 0 + an$,which implies $

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$\frac{v(6n - 9)}{2n}$

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(A) Given that the body starts from rest,initial velocity $u = 0$. Using the first equation of motion,$v = u + at$,we get $v = 0 + an$,which implies $a = \frac{v}{n}$. The total displacement in $n \ s$ is $S_n = \frac{1}{2}an^2$. The displacement in the first $(n-3) \ s$ is $S_{n-3} = \frac{1}{2}a(n-3)^2$. The displacement in the last $3 \ s$ is given by $\Delta S = S_n - S_{n-3}$. Substituting the values: $\Delta S = \frac{1}{2}a[n^2 - (n-3)^2]$. Expanding the term: $\Delta S = \frac{1}{2}a[n^2 - (n^2 - 6n + 9)] = \frac{1}{2}a(6n - 9)$. Substituting $a = \frac{v}{n}$: $\Delta S = \frac{1}{2} \cdot \frac{v}{n} \cdot (6n - 9) = \frac{v(6n - 9)}{2n}$.

$A$ body starts from rest with uniform acceleration $a$. Its velocity after $n \ s$ is $v$. The displacement of the body in the last $3 \ s$ is: (Assume total time of journey is $n \ s$)

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