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 .

Physics · Motion in one Dimension

JEE Main 2024 — 30 January, Shift 1 — Question 51

The displacement and the increase in the velocity of a moving particle in the time interval of t to $(\mathrm{t}+1) \mathrm{s}$ are 125 m and $50 \mathrm{~m} / \mathrm{s}$ , respectively. The distance travelled by the particle in $(t+2)^{\mathrm{th}} \mathrm{s}$ is $\_\_\_\_\_\_\_$ m .

Answer: 175

Numerical answer — enter this value.

Step-by-step solution

Considering acceleration is constant $v=u+a t$ $u+50=u+a \Rightarrow a=50 \mathrm{~m} / \mathrm{s}^{2}$

$125=u t+\frac{1}{2} a t^{2}$

$125=u+\frac{a}{2}$

$\Rightarrow u=100 \mathrm{~m} / \mathrm{s}$

$\therefore S_{n^{t h}}=u+\frac{a}{2}[2 n-1]$

$=175 \mathrm{~m}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 30 January, Shift 1
Subject: Physics
Chapter: Motion in one Dimension
Topic: Uniformly Accelerated Motion