A body A starts from rest with an acceleratio | vedclass

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(A) The distance traveled by an object in the $n$th second is given by $S_n = u + \frac{a}{2}(2n - 1)$.For body $A$,the distance traveled in the $5$th

Vedclass · Accepted Answer

$5:9$

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(A) The distance traveled by an object in the $n$th second is given by $S_n = u + \frac{a}{2}(2n - 1)$.For body $A$,the distance traveled in the $5$th second is $S_{A,5} = 0 + \frac{a_1}{2}(2 	imes 5 - 1) = \frac{9a_1}{2}$.Body $B$ starts $2$ seconds after $A$. Therefore,the $5$th second after the start of $A$ corresponds to the $(5 - 2) = 3$rd second of motion for body $B$.The distance traveled by body $B$ in its $3$rd second is $S_{B,3} = 0 + \frac{a_2}{2}(2 	imes 3 - 1) = \frac{5a_2}{2}$.Given that these distances are equal: $\frac{9a_1}{2} = \frac{5a_2}{2}$.Simplifying this,we get $9a_1 = 5a_2$,which implies $\frac{a_1}{a_2} = \frac{5}{9}$.

$A$ body $A$ starts from rest with an acceleration $a_1$. After $2$ seconds,another body $B$ starts from rest with an acceleration $a_2$. If they travel equal distances in the $5$th second after the start of $A$,then the ratio $a_1:a_2$ is equal to

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