Question 130: Motion in 1D

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:

Accepted Answer

To find the ratio of the initial velocity to the initial acceleration of a particle moving along the $x$-axis, we need to differentiate the given position function with respect to time $t$.
Let's start by writing the position function:
$x = \alpha t^4 + \beta t^2 + \gamma t + \delta$
To find the velocity ($v$), we differentiate $x$ with respect to $t$:
$v = \frac{dx}{dt} = \frac{d}{dt} (\alpha t^4 + \beta t^2 + \gamma t + \delta)$
This gives us:
$v = 4\alpha t^3 + 2\beta t + \gamma$
The initial velocity is the velocity at $t = 0$:
$v(0) = 4\alpha (0)^3 + 2\beta (0) + \gamma = \gamma$
Next, to find the acceleration ($a$), we differentiate $v$ with respect to $t$:
$a = \frac{dv}{dt} = \frac{d}{dt} (4\alpha t^3 + 2\beta t + \gamma)$
This gives us:
$a = 12\alpha t^2 + 2\beta$
The initial acceleration is the acceleration at $t = 0$:
$a(0) = 12\alpha (0)^2 + 2\beta = 2\beta$
Now, we find the ratio of the initial velocity to the initial acceleration:
$	ext{Ratio} = \frac{v(0)}{a(0)} = \frac{\gamma}{2\beta}$
Therefore, the correct answer is:
Option D: $\gamma: 2\beta$

Answer

$2 \alpha: \delta$

Answer

$\gamma: 2 \delta$

Answer

$4 \alpha: \beta$

Answer

$\gamma: 2 \beta$

Motion in 1D

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:

Choose the correct answer: