Q 11-02-023NEETJEE MainMedium
A body moving with uniform acceleration covers $12$ m in the $3^{\text{rd}}$ second and $20$ m in the $5^{\text{th}}$ second of its motion. Its initial velocity and acceleration are respectively
Answer: (D) $2\ \text{m/s},\ 4\ \text{m/s}^2$
Distance in the $n^{\text{th}}$ second: $s_n = u + \dfrac{a}{2}(2n - 1)$.
$$s_3 = u + 2.5a = 12, \qquad s_5 = u + 4.5a = 20$$
Subtracting: $2a = 8$, so $a = 4\ \text{m/s}^2$. Then $u = 12 - 10 = 2$ m/s.
Solution by Sreeraj P, M.Sc Physics