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BITSAT 2024Paper 1
Exam
BITSAT
Year
2024
Paper
Paper 1
Subject
Physics
Chapter
Motion in One Dimension

BITSAT 2024 · Paper 1 · Physics · Motion in One Dimension

The distance travelled by a particle starting from rest and moving...

The distance travelled by a particle starting from rest and moving with an acceleration 43ms−2\frac{4}{3} \mathrm{ ms}^{-2}, in the third second is:

Options

  1. A6 m
  2. B4 m
  3. C103m\frac{10}{3} \mathrm{ m}
  4. D193m\frac{19}{3} \mathrm{ m}

Correct answer

C. 103m\frac{10}{3} \mathrm{ m}

Step-by-step solution

Distance travelled in the nth second is given by tn=u+a2(2n−1)\mathrm{t}_{\mathrm{n}}=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \mathrm{n}-1) put u=0,a=43ms−2,n=3∴d=0+43×2(2×3−1)=46×5=103m\mathrm{u}=0, \mathrm{a}=\frac{4}{3} \mathrm{ ms}^{-2}, \mathrm{n}=3\therefore \quad d=0+\frac{4}{3 \times 2}(2 \times 3-1)=\frac{4}{6} \times 5=\frac{10}{3} m
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