Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

The distance travelled by a particle starting from rest and moving with an acceleration 4/3 ms⁻², in the third second is

6 m

4 m

10/3 m

19/3 m

Identify Given Values: Initial velocity ( $u$ ) = 0 (starts from rest). Acceleration ( $a$ ) = $4/3 \text{ ms}^{-2}$ .

Select Formula: The distance travelled in the $n$ -th second ( $S_n$ ) for uniformly accelerated motion is given by: $S_n = u + \frac{a}{2}(2n - 1)$
Calculation: Substitute the values into the formula: $S_{3rd} = 0 + \frac{4/3}{2} (2 \times 3 - 1)$ $S_{3rd} = \frac{2}{3} (6 - 1)$ $S_{3rd} = \frac{2}{3} \times 5 = \frac{10}{3} \text{ m}$

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