If a body starts from rest and travels 120 , | vedclass

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(D) The formula for distance traveled in the $n^{th}$ second is given by $S_n = u + \frac{a}{2}(2n - 1)$.Given that the body starts from rest,the init

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$21.8 \, cm/s^2$

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(D) The formula for distance traveled in the $n^{th}$ second is given by $S_n = u + \frac{a}{2}(2n - 1)$.Given that the body starts from rest,the initial velocity $u = 0 \, m/s$.The distance traveled in the $6^{th}$ second is $S_6 = 120 \, cm = 1.2 \, m$.Substituting the values into the formula:$1.2 = 0 + \frac{a}{2}(2 	imes 6 - 1)$$1.2 = \frac{a}{2}(11)$$a = \frac{1.2 	imes 2}{11} = \frac{2.4}{11} \, m/s^2 \approx 0.218 \, m/s^2$.Converting back to $cm/s^2$:$a = 0.218 	imes 100 \, cm/s^2 = 21.8 \, cm/s^2$.

If a body starts from rest and travels $120 \, cm$ in the $6^{th}$ second,then the acceleration of the body is:

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