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

1. Velocity (v) 

Given position $x$ as a function of time, velocity is:

$$ v = \frac{dx}{dt} = 4\alpha t^3 + 2\beta t + \gamma $$

Initial velocity at $t = 0$:

$$ v(0) = 4\alpha(0)^3 + 2\beta(0) + \gamma = \gamma $$

$$ v_{\text{initial}} = \gamma $$

2. Acceleration (a)

Acceleration is the derivative of velocity:

$$ a = \frac{dv}{dt} = \frac{d}{dt}(4\alpha t^3 + 2\beta t + \gamma) $$ $$ a = 12\alpha t^2 + 2\beta $$

Initial acceleration at $t = 0$:

$$ a(0) = 12\alpha(0)^2 + 2\beta = 2\beta $$

$$ a_{\text{initial}} = 2\beta $$

3. Ratio of initial velocity to initial acceleration

$$ \text{Ratio} = \frac{v_{\text{initial}}}{a_{\text{initial}}} = \frac{\gamma}{2\beta} $$

$$ \frac{v_{\text{initial}}}{a_{\text{initial}}} = \frac{\gamma}{2\beta} $$