Q73 / 100 · BITSAT 2024 (Memory Based Paper 1) · Physics

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

  • A
    6 m
  • B
    4 m
  • Correct answer
    C
    $\frac{10}{3} \mathrm{~m}$
  • D
    $\frac{19}{3} \mathrm{~m}$

Solution

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

Hence, the correct answer is option: 'C'.

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