A body starts moving with an initial velocity of $4 m s^{- 1}$ and acceleration of $x m s^{- 2 .}$ If the distance traveled by it is $30 m$ in $2^{nd} s$ , then the value of $x$ is ________.

$4$

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$30$

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$10.3$

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$17.33$

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Solution

The correct option is D
$17.33$

Step 1: Given data and formula
$Initial velocity u = 4 {ms}^{- 1}, n = 2,$
Distance traveled during nth second $S_{n} = 30$
Formula used :
Distance traveled during nth second $S_{n} = u + a (n - \frac{1}{2})$
Step 2: Calculation of acceleration $(a)$
On putting given value in above formula
$30 = 4 + a (2 - \frac{1}{2}) 30 - 4 = a \times \frac{3}{2} a = \frac{26 \times 2}{3} = 17.33 {ms}^{- 2}$
Hence, the correct option is (d).

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