Q122 / 160 · AP EAMCET 2024 (20 May Shift 1) · Physics

A body starting from rest moving with an acceleration of $\frac{5}{4} \mathrm{~ms}^{-2}$. The distance travelled by the body in the third second is:

  • A
    $\frac{15}{8} m$
  • Correct answer
    B
    $\frac{25}{8} m$
  • C
    $\frac{25}{4} m$
  • D
    $\frac{12}{7} m$

Solution

$\mathrm{u}=0, \mathrm{a}=\frac{5}{4} \mathrm{~ms}^{-2}$ $S_{n t h}=u+(2 n-1) \frac{a}{2}$ $\therefore \quad \mathrm{S}_{3 \mathrm{rd}}=0+(2 \times 3-1) \times \frac{5}{8}=\frac{25}{8} \mathrm{~m}$

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

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