Q 11-02-067JEE MainJEE Main 2024 (9 Apr, Shift 1)Medium
A particle moving in a straight line covers half the distance with speed $6\ \text{m/s}$. The other half is covered in two equal time intervals with speeds $9\ \text{m/s}$ and $15\ \text{m/s}$ respectively. The average speed of the particle during the motion is:
Answer: (B) $8\ \text{m/s}$
Let the total distance be $2s$. First half: $t_1 = \dfrac s6$.
Second half in two equal intervals $t$: $9t + 15t = s \Rightarrow t = \dfrac{s}{24}$, so $t_2 = 2t = \dfrac{s}{12}$.
$$v_{avg} = \frac{2s}{\frac s6 + \frac s{12}} = \frac{2s}{s/4} = 8\ \text{m/s}$$
Solution by Sreeraj P, M.Sc Physics