Question 125: 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: