Q110 / 200 · NEET 2024 (Re-NEET) · Physics

A particle is moving along $x$-axis with its position (x) varying with time $(t)$ as $x=\alpha t^4+\beta t^2+\gamma t+\delta$. The ratio of its initial velocity to its initial acceleration, respectively, is:

  • A
    $2 \alpha: \delta$
  • B
    $\gamma: 2 \delta$
  • C
    $4 \alpha: \beta$
  • Correct answer
    D
    $\gamma: 2 \beta$

Solution

Position of particle, $x=\alpha t^4+\beta t^2+\gamma t+\delta$ Velocity $v=\frac{d x}{d t}=4 \alpha t^3+2 \beta t+\gamma$ Initial velocity $=v(t=0)=\gamma$ Acceleration $a=\frac{d v}{d t}=12 \alpha t^2+2 \beta$ Initial acceleration $=a(t=0)=2 \beta$ $\therefore \frac{v(t=0)}{a(t=0)}=\frac{\gamma}{2 \beta}$

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

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