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Current Question (ID: 8836)

Question:

$\text{The motion of a particle is given by the equation } S = (3t^3 + 7t^2 + 14t + 8) \text{ m}. \text{ The value of the acceleration of the particle at } t = 1 \text{ is:}$

Options:

1. $10 \text{ m/s}^2$
2. $32 \text{ m/s}^2$
3. $23 \text{ m/s}^2$
4. $16 \text{ m/s}^2$

Solution:

$\text{Hint: Velocity of the particle is given by } v = \frac{ds}{dt}$ $\text{Step: Find the acceleration of the particle.}$ $\text{First, find velocity by differentiating the position equation:}$ $v = \frac{ds}{dt} = \frac{d}{dt}(3t^3 + 7t^2 + 14t + 8)$ $= 9t^2 + 14t + 14$ $\text{Then, find acceleration by differentiating velocity:}$ $a = \frac{dv}{dt} = 18t + 14$ $\text{At } t = 1 \text{ sec:}$ $a = 18(1) + 14 = 32 \text{ m/s}^2$ $\text{Hence, option (2) is the correct answer.}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}