A body starts from rest with an acceleration | vedclass

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Question

The distance covered in fifth second for first body is equal to the distance travelled in $3^{	ext {rd }}$ second for second body, $S _{5}= S _{3}$

Vedclass · Accepted Answer

$5: 9$

Vedclass · Answer

The distance covered in fifth second for first body is equal to the distance travelled in $3^{	ext {rd }}$ second for second body, $S _{5}= S _{3}$
$S_{t}= u +\frac{(2 t -1) a}{2}$
$S_{5}=0+\frac{(2 	imes 5-1) a}{2}$
$S_{5}=\frac{9 a_{1}}{2}$
$S_{3}=0+\frac{5}{2} a_{2}=\frac{5}{2} a_{2}$
$\frac{9 a_{1}}{2}=\frac{5}{2} a_{2}$
$\frac{a_{1}}{a_{2}}=\frac{5}{9}$

A body 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 distance in fifth second, after the starts of $A$, the ratio $a _{1}: a _{2}$ will be equal to

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