Q 11-02-109JEE MainJEE Main 2021 (26 Feb, Shift 2)Easy
A scooter accelerates from rest for time $t_1$ at constant rate $a_1$ and then retards at constant rate $a_2$ for time $t_2$ and comes to rest. The correct value of $\frac{t_1}{t_2}$ will be:
Answer: (A) $\frac{a_2}{a_1}$
The maximum speed reached is $v = a_1t_1$, and it falls to zero in time $t_2$, so $v = a_2t_2$.
$$a_1t_1 = a_2t_2 \Rightarrow \frac{t_1}{t_2} = \frac{a_2}{a_1}$$
Solution by Sreeraj P, M.Sc Physics