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(C) The distance covered by a body in the $n^{	ext{th}}$ second is given by $S_{n} = u + \frac{a}{2}(2n - 1)$.For body $A$,starting from rest $(u=0)$

Vedclass · Accepted Answer

$5: 9$

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(C) The distance covered by a body in the $n^{	ext{th}}$ second is given by $S_{n} = u + \frac{a}{2}(2n - 1)$.For body $A$,starting from rest $(u=0)$ at $t=0$,the distance covered in the $5^{	ext{th}}$ 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 $5^{	ext{th}}$ second after the start of $A$ corresponds to the $(5-2) = 3^{	ext{rd}}$ second of motion for body $B$.For body $B$,starting from rest $(u=0)$,the distance covered in its $3^{	ext{rd}}$ second is:$S_{B, 3} = 0 + \frac{a_{2}}{2}(2 	imes 3 - 1) = \frac{5a_{2}}{2}$.Given that $S_{A, 5} = S_{B, 3}$,we have:$\frac{9a_{1}}{2} = \frac{5a_{2}}{2}$.Thus,$\frac{a_{1}}{a_{2}} = \frac{5}{9}$.

$A$ body $A$ starts from rest with an acceleration $a_{1}$. After two seconds,another body $B$ starts from rest with an acceleration $a_{2}$. If they travel equal distances in the fifth second after the start of $A$,the ratio $a_{1}: a_{2}$ will be equal to:

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