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Current Question (ID: 8930)

Question:

$\text{Two cars } A \text{ and } B \text{ are travelling in the same direction with velocities } v_1 \text{ and } v_2 (v_1 > v_2)\text{. When the car } A \text{ is at a distance } d \text{ behind the car } B\text{, the driver of the car } A \text{ applied the brake producing uniform retardation } a\text{. There will be no collision when:}$

Options:

1. $d < \frac{(v_1-v_2)^2}{2a}$
2. $d < \frac{v_1^2-v_2^2}{2a}$
3. $d > \frac{(v_1-v_2)^2}{2a}$
4. $d > \frac{v_1^2-v_2^2}{2a}$

Solution:

$\text{Hint: Use the concept of relative velocity.}$ $\text{Step: Find the condition of the actual distance between two cars.}$ $\text{The initial relative velocity between the cars is given by: } v_1 - v_2$ $\text{The final relative velocity between the cars will be 0.}$ $\text{From the third equation of motion is given by: } v^2 = u^2 - 2as$ $\Rightarrow 0 = (v_1 - v_2)^2 - 2 \times a \times s$ $\Rightarrow s = \frac{(v_1-v_2)^2}{2a}$ $\text{If the actual distance between two cars is } d \text{ then the distance between two cars is } s \text{ then a collision will take place. To avoid the collision, } d > s \text{ i.e., } d > \frac{(v_1-v_2)^2}{2a}$ $\text{Hence, option (3) is the correct answer.}$

Import JSON File

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
}