A body A starts from rest with an acceleratio | vedclass

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(A) The distance traveled by a body in the $n^{th}$ second is given by $S_n = u + \frac{a}{2}(2n - 1)$.For body $A$,it starts from rest $(u=0)$ and we

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$5 : 9$

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(A) The distance traveled by a body in the $n^{th}$ second is given by $S_n = u + \frac{a}{2}(2n - 1)$.For body $A$,it starts from rest $(u=0)$ and we consider the $5^{th}$ second of its motion. Thus,$S_{A} = 0 + \frac{a_1}{2}(2 	imes 5 - 1) = \frac{9a_1}{2}$.Body $B$ starts $2 \ s$ after $A$. Therefore,the $5^{th}$ second after the start of $A$ corresponds to the $(5-2) = 3^{rd}$ second of the motion of body $B$.For body $B$,$S_{B} = 0 + \frac{a_2}{2}(2 	imes 3 - 1) = \frac{5a_2}{2}$.Given that the distances are equal,$S_A = S_B$.$\frac{9a_1}{2} = \frac{5a_2}{2}$.$9a_1 = 5a_2 \Rightarrow \frac{a_1}{a_2} = \frac{5}{9}$.

$A$ body $A$ starts from rest with an acceleration $a_1$. After $2 \ s$,another body $B$ starts from rest with an acceleration $a_2$. If they travel equal distance in the $5^{th} \ s$ after the start of $A$,then the ratio $a_1 : a_2$ is equal to

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