A body $A$ start from rest with an acceleration $a_{1}$ . After $2$ seconds, another body $B$ start from rest with an acceleration $a_{2}$ . if they travel equal distance in the $5 t h$ second, after the start of A, then the ratio $a_{1} : a_{2}$

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Solution

We have,

$s = u + \frac{a}{2} (2 n - 1)$

$\to u = 0$

$∴ \frac{a_{1}}{2} (2 (5) - 1) = \frac{a_{2}}{2} (2 (3) - 1)$

$9 a_{1} = 5 a_{2}$

$\frac{a_{1}}{a_{2}} = \frac{5}{9}$

$∴ a_{1} : a_{2} = 5 : 9$

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A body A start from rest with an acceleration a1. After 2 seconds, another body B start from rest with an acceleration a2. if they travel equal distance in the 5th second, after the start of A, then the ratio a1:a2

We have,s=u+a2(2n−1)→u=0∴a12(2(5)−1)=a22(2(3)−1)9a1=5a2a1a2=59∴a1:a2=5:9

A body $A$ start from rest with an acceleration $a_{1}$ . After $2$ seconds, another body $B$ start from rest with an acceleration $a_{2}$ . if they travel equal distance in the $5 t h$ second, after the start of A, then the ratio $a_{1} : a_{2}$

We have,

$s = u + \frac{a}{2} (2 n - 1)$

$\to u = 0$

$∴ \frac{a_{1}}{2} (2 (5) - 1) = \frac{a_{2}}{2} (2 (3) - 1)$

$9 a_{1} = 5 a_{2}$

$\frac{a_{1}}{a_{2}} = \frac{5}{9}$

$∴ a_{1} : a_{2} = 5 : 9$