A stone falls freely from rest and the total distance covered in the last second equals the distance covered in the first 3 seconds (g = 10text m/s ^2). The time of flight is 5 s. Using the same setup: the distance covered by a particle starting from rest with a = dfrac 4 3 text m/s ^2 in the third second of motion is:

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NEET Physics QUESTION #7309

Question 1

A stone falls freely from rest and the total distance covered in the last second equals the distance covered in the first 3 seconds ($g = 10\text{ m/s}^2$). The time of flight is 5 s. Using the same setup: the distance covered by a particle starting from rest with $a = \dfrac{4}{3}\text{ m/s}^2$ 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

Distance in $n$-th second from rest: $s_n = \dfrac{a}{2}(2n-1)$

For $n = 3$, $a = \dfrac{4}{3}\text{ m/s}^2$:

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

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