Q 11-02-069JEE MainJEE Main 2024 (30 Jan, Shift 1)Medium
The displacement and the increase in the velocity of a moving particle in the time interval of $t$ to $(t+1)\ \text{s}$ are $125\ \text{m}$ and $50\ \text{m s}^{-1}$, respectively. The distance travelled by the particle in the $(t+2)^{\text{th}}$ second is ______ m.
Numerical value type. Enter your answer.
Answer: 175
The velocity increases by $50\ \text{m s}^{-1}$ in 1 s, so $a = 50\ \text{m s}^{-2}$.
In the interval $t$ to $t+1$: $125 = v_t(1) + \tfrac12(50)(1)^2 \Rightarrow v_t = 100\ \text{m s}^{-1}$.
The $(t+2)^{\text{th}}$ second is the interval $t+1$ to $t+2$, which starts with $v = 150\ \text{m s}^{-1}$:
$$s = 150 + \tfrac12(50) = 175\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics