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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:
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The correct answer is:
$\gamma: 2 \beta$
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 $=a(t=0)=2 \beta$
$\therefore \frac{v(t=0)}{a(t=0)}=\frac{\gamma}{2 \beta}$
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 $=a(t=0)=2 \beta$
$\therefore \frac{v(t=0)}{a(t=0)}=\frac{\gamma}{2 \beta}$
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