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 __________ \(\mathrm{m}\).
Pick an option to check your answer.
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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