Q 11-02-025JEE MainHard
Car A starts from rest with uniform acceleration $a_1$. Three seconds later, car B starts from rest from the same point with uniform acceleration $a_2$. If the displacement of A in its $6^{\text{th}}$ second equals the displacement of B in the same interval of time, then $a_1 : a_2$ is
Answer: (B) $5 : 11$
From rest, the distance in the $n^{\text{th}}$ second is $s_n = \dfrac{a}{2}(2n - 1)$.
A's $6^{\text{th}}$ second is the interval $t = 5$ s to $6$ s. B started $3$ s later, so this is B's $3^{\text{rd}}$ second.
$$\frac{a_1}{2}(2 \times 6 - 1) = \frac{a_2}{2}(2 \times 3 - 1) \;\Rightarrow\; 11a_1 = 5a_2$$
$$a_1 : a_2 = 5 : 11$$
Solution by Sreeraj P, M.Sc Physics