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JEE Mains · Physics · STD 11 - Motion in straight line
  • JEE Mains 2024
  • Hard
  • Numerical

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

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Verified Solution

Answer & Solution

Correct Answer

(B) \(175\)

Step-by-step Solution

Detailed explanation

Considering acceleration is constant

\( v=u+a t \)

\( u+50=u+a \Rightarrow a=50 \mathrm{~m} / \mathrm{s}^2 \)

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

\( 125=u+\frac{a}{2} \)

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

\( \therefore S_{n^{r k}}=u+\frac{a}{2}[2 n-1] \)

\( =175 \mathrm{~m} \)

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