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Current Question (ID: 8834)

Question:

$\text{A particle moves along a straight line } OX. \text{ At a time } t \text{ (in seconds), the displacement } x \text{ (in metres) of the particle from } O \text{ is given by } x = 40 + 12t - t^3. \text{ How long would the particle travel before coming to rest?}$

Options:

1. $24 \text{ m}$
2. $40 \text{ m}$
3. $56 \text{ m}$
4. $16 \text{ m}$

Solution:

$\text{Hint: When the particle stops, the velocity of the particle is equal to zero.}$ $\text{Step 1: Find the instantaneous velocity of the particle.}$ $\text{Displacement of the particle is, } x = 40 + 12t - t^3$ $\text{The velocity is given by, } v = \frac{dx}{dt}$ $\therefore v = \frac{d}{dt}(40 + 12t - t^3) = 0 + 12 - 3t^2$ $\text{Step 2: Put the velocity equal to zero.}$ $\text{When the final velocity, } v = 0,$ $\therefore 12t - 3t^2 = 0$ $\text{Or } t^2 = \frac{12}{3} = 4$ $\text{Or } t = 2\text{s}$ $\text{Step 3: Find the distance travelled by the particle.}$ $\text{The initial position of the particle, } x_i = 40 \text{ m}$ $\text{The final position of the particle,}$ $x_f = 40 + 12(2) - (2)^3$ $= 40 + 24 - 8 = 64 - 8$ $= 56 \text{ m}$ $\text{Hence, the distance covered by the particle before coming to stop will be } 56 - 40 = 16 \text{ m}.$ $\text{Hence, option (4) is the correct answer.}$

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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
}