The displacement and the increase in the velocity of a moving particle in the time interval of t to (t+1) s are 125 m and 50 m/s, respectively. The distance travelled by the particle in (t+2)th s is _________ m. Numerical answer type. Work it out, then open the answer.

Answer: 175. Increase in velocity in 1 s is 50 m/s, so a = 50 m/s². Displacement in t to t+1: 125 = vt(1) + 1/2(50)(1)² ⇒ vt = 100 m/s. The (t+2) th second is from t+1 to t+2, starting at vt+1 = 150 m/s. s = 150(1) + 1/2(50)(1)² = 175 m

The displacement and the increase in the velocity of a moving… | JEE Main 2024 (30 Jan, Shift 1)

Q11JEE Main 2024 · 30 JAN · SHIFT 1Medium

Equations of Kinematics

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 $50\text{ m/s}$ , respectively. The distance travelled by the particle in $(t+2)^{\text{th}}\text{ s}$ is _________ m.

Numerical answer type. Work it out, then open the answer.

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Answer: 175

Increase in velocity in 1 s is 50 m/s, so $a = 50\text{ m/s}^2$ .
Displacement in $t$ to $t+1$ : $125 = v_t(1) + \frac{1}{2}(50)(1)^2 \Rightarrow v_t = 100\text{ m/s}$ .
The $(t+2)$ th second is from $t+1$ to $t+2$ , starting at $v_{t+1} = 150\text{ m/s}$ .
$s = 150(1) + \frac{1}{2}(50)(1)^2 = 175\text{ m}$

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