Understanding Particle Motion: Position, Velocity, and Acceleration

This problem asks us to analyze the motion of a particle based on its position given as a function of time. Specifically, we need to determine the nature of its acceleration. The position ($x$) is provided by the equation $x(t) = At^3$, where $A$ is a constant that is not zero.

Key Concepts:

• Position: The location of a particle at a specific time ($x(t)$).
• Velocity: The rate at which the position changes over time ($v(t)$). It's the first derivative of position with respect to time.
• Acceleration: The rate at which the velocity changes over time ($a(t)$). It's the first derivative of velocity or the second derivative of position with respect to time.

Deriving Acceleration from the Position Equation

We are given the position equation $x(t) = At^3$. To find the acceleration, we will perform differentiation twice.

Step 1: Find the Velocity ($v(t)$)

Velocity is the rate of change of position, so we differentiate $x(t)$ with respect to time $t$. We use the power rule for differentiation, which states that $\frac{d}{dt}(t^n) = nt^{n-1}$.

$v(t) = \frac{dx}{dt}$

Substitute the given position equation:

$v(t) = \frac{d}{dt}(At^3)$

Since $A$ is a constant, we can take it out of the differentiation:

$v(t) = A \cdot \frac{d}{dt}(t^3)$

Applying the power rule with $n=3$:

$v(t) = A \cdot (3t^{3-1})$

$v(t) = 3At^2$

This shows that the velocity of the particle is directly proportional to the square of time ($t^2$).

Step 2: Find the Acceleration ($a(t)$)

Acceleration is the rate of change of velocity, so we differentiate $v(t)$ with respect to time $t$. We will again use the power rule.

$a(t) = \frac{dv}{dt}$

Substitute the derived velocity equation:

$a(t) = \frac{d}{dt}(3At^2)$

Since $3A$ is a constant, we take it out of the differentiation:

$a(t) = 3A \cdot \frac{d}{dt}(t^2)$

Applying the power rule with $n=2$:

$a(t) = 3A \cdot (2t^{2-1})$

$a(t) = 6At$

Nature of the Acceleration

Our calculation shows that the acceleration of the particle is given by the equation $a(t) = 6At$. We are given that $A$ is a non-zero constant.

Let's examine the relationship between acceleration ($a(t)$) and time ($t$):

$a(t) = (6A) \cdot t$

Since $6A$ is a constant value (because $A$ is a constant), the equation $a(t) = 6At$ indicates that the acceleration is directly proportional to the time $t$. As time $t$ increases, the acceleration increases linearly.

Evaluating the Options Provided:

• Option 1: The acceleration is constant. This is incorrect because $a(t) = 6At$ depends on $t$, so it changes with time. A constant acceleration would mean $a(t)$ equals a fixed number, not dependent on $t$.
• Option 2: The acceleration is directly proportional to time $t$. This matches our result $a(t) = 6At$. The acceleration varies linearly with time.
• Option 3: The acceleration is directly proportional to the square of time $t^2$. This is incorrect. Our derived acceleration is proportional to $t$, not $t^2$. (Note: The velocity $v(t) = 3At^2$ is proportional to $t^2$).
• Option 4: The acceleration is zero. This is incorrect. Since $A$ is non-zero, and for any time $t \neq 0$, the acceleration $a(t) = 6At$ will be non-zero.

Final Conclusion

By differentiating the given position equation $x(t) = At^3$ twice with respect to time, we found the acceleration to be $a(t) = 6At$. This mathematical form clearly shows that the acceleration is directly proportional to time $t$.