Question

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. \(4\alpha : \beta\)
• D. \(\gamma : 2\beta\)

Answer 1

The Correct Option is D

Velocity (v): Defined as the rate of change of displacement with respect to time, $v = \frac{dx}{dt} = 4\alpha t^3 + 2\beta t + \gamma$. The initial velocity, at time $t = 0$, is $\gamma$.

Acceleration (a): Defined as the rate of change of velocity with respect to time, $a = \frac{dv}{dt} = 12\alpha t^2 + 2\beta$. The initial acceleration, at time $t = 0$, is $2\beta$.

Ratio of Initial Velocity to Initial Acceleration: This ratio is calculated as $\frac{\gamma}{2\beta}$.