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Q.
If a body starts from rest and travels $110\, cm$ in the $9^{th}$ second, then acceleration of the body is:
Jharkhand CECEJharkhand CECE 2003
Solution:
The distance travelled by the body in nth second is given by
$s_{n}=u+\frac{1}{2} a(2 n-1)$..(i)
Since, body starts from rest, so $u=0$.
Given, $s_{n}=110 \,cm , n=9$
Substituting the values in Eq. (i), we get
$110=0+\frac{1}{2} a(2 \times 9-1)$
or $110=\frac{a}{2} \times 17$
$\therefore a=\frac{2 \times 110}{17} \approx 13\, cm / s ^{2}$
or $ a=0.13 \,m / s ^{2}$