A body starts from rest under uniform acceleration. If distance covered by it in $3^{r d}$ second is $3 m$ , then the distance covered in $5^{t h}$ second will be

5 m

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3 m

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\frac{27}{5} m

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\frac{24}{5} m

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Solution

The correct option is B $\frac{27}{5} m$
$s = u + \frac{a}{2} (2 n - 1)$
As, $3 = 0 + \frac{a}{2} [2 \times 3 - 1]$
$∴ a = \frac{6}{5}$
Now, $s^{'} = 0 + \frac{6}{2 \times 5} (2 \times 5 - 1)$
$= \frac{6}{2 \times 5} \times 9 = \frac{54}{10}$
$= \frac{27}{5} m$

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A body starts from rest under uniform acceleration. If distance covered by it in 3rd second is 3m, then the distance covered in 5th second will be

The correct option is B 275ms=u+a2(2n−1) As, 3=0+a2[2×3−1]∴a=65Now,s′=0+62×5(2×5−1) =62×5×9=5410=275m

A body starts from rest under uniform acceleration. If distance covered by it in $3^{r d}$ second is $3 m$ , then the distance covered in $5^{t h}$ second will be

The correct option is B $\frac{27}{5} m$
$s = u + \frac{a}{2} (2 n - 1)$
As, $3 = 0 + \frac{a}{2} [2 \times 3 - 1]$
$∴ a = \frac{6}{5}$
Now, $s^{'} = 0 + \frac{6}{2 \times 5} (2 \times 5 - 1)$
$= \frac{6}{2 \times 5} \times 9 = \frac{54}{10}$
$= \frac{27}{5} m$