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Current Question (ID: 19024)

Question:

$\text{The position of a particle as a function of time } t, \text{ is given by:}$ $x(t) = at + bt^2 - ct^3$ $\text{where } a, b \text{ and } c \text{ are constants. When the particle attains zero acceleration, then its velocity will be:}$

Options:

1. $a + \frac{b^2}{4c}$
2. $a + \frac{b^2}{c}$
3. $a + \frac{b^2}{3c}$
4. $a + \frac{b^2}{2c}$

Solution:

$\text{Hint: } a = \frac{d^2x}{dt^2}$ $x = at + bt^2 - ct^3$ $v = a + 2bt - 3ct^2$ $\text{acceleration} = 2b - 6ct = 0$ $t = \frac{b}{3c}$ $\text{so velocity at } t = \frac{b}{3c}$ $v = a + 2b \cdot \frac{b}{3c} - 3c \cdot \left(\frac{b}{3c}\right)^2$ $= a + \frac{b^2}{3c}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}