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JEE Mains · Physics · STD 11 - 2. motion in straight line

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 __________ \(\mathrm{m}\).

  1. A \(24\)
  2. B \(175\)
  3. C \(458\)
  4. D \(157\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(175\)

Step-by-step Solution

Detailed explanation

Considering acceleration is constant \( v=u+a t \) \( u+50=u+a \Rightarrow a=50 \mathrm{~m} / \mathrm{s}^2 \) \( 125=u t+\frac{1}{2} a t^2 \) \( 125=u+\frac{a}{2} \) \( \Rightarrow u=100 \mathrm{~m} / \mathrm{s} \) \( \therefore S_{n^{r k}}=u+\frac{a}{2}[2 n-1] \)…
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