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NEET Physics QUESTION #7310
Question 1

A particle starting from rest moves with acceleration $\dfrac{4}{3}\text{ m/s}^2$. The distance it travels in the third second of motion is:

  • $\dfrac{10}{3}\text{ m}$✔️
  • $\dfrac{19}{3}\text{ m}$
  • $6\text{ m}$
  • $4\text{ m}$
Correct Answer Explanation

Using the formula for distance in the $n$-th second: $s_n = u + \dfrac{a}{2}(2n-1)$, with $u = 0$, $a = \dfrac{4}{3}$, $n = 3$:

$s_3 = 0 + \dfrac{4/3}{2}\times(2\times3-1) = \dfrac{2}{3}\times 5 = \mathbf{\dfrac{10}{3}\text{ m}}$