Understanding Particle Motion: Distance, Velocity, and Acceleration

The question asks for the acceleration of a particle given its distance traveled along a straight line as a function of time. The distance is given by the equation $x = t^3 (t - 6)$, where $x$ is the distance and $t$ is the time in seconds.

In kinematics, which is the study of motion without considering the forces that cause the motion, the relationship between distance, velocity, and acceleration is defined through differentiation with respect to time.

• Velocity ($v$) is the rate of change of distance with respect to time. Mathematically, this is the first derivative of the distance function $x(t)$ with respect to time $t$: $v(t) = \frac{dx}{dt}$.
• Acceleration ($a$) is the rate of change of velocity with respect to time. Mathematically, this is the first derivative of the velocity function $v(t)$ with respect to time $t$, or the second derivative of the distance function $x(t)$ with respect to time $t$: $a(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2}$.

Step-by-Step Calculation of Acceleration

First, let's simplify the given equation for the distance $x$:

$x = t^3 (t - 6)$

Distribute $t^3$ into the parenthesis:

$x = t^3 \cdot t - t^3 \cdot 6$

$x = t^{3+1} - 6t^3$

$x = t^4 - 6t^3$

Now, we find the velocity $v(t)$ by differentiating $x(t)$ with respect to $t$. We use the power rule for differentiation, which states that $\frac{d}{dt}(t^n) = nt^{n-1}$.

$v(t) = \frac{dx}{dt} = \frac{d}{dt}(t^4 - 6t^3)$

$v(t) = \frac{d}{dt}(t^4) - \frac{d}{dt}(6t^3)$

$v(t) = (4 \cdot t^{4-1}) - (6 \cdot 3 \cdot t^{3-1})$

$v(t) = 4t^3 - 18t^2$

So, the velocity of the particle is given by $v(t) = 4t^3 - 18t^2$.

Next, we find the acceleration $a(t)$ by differentiating the velocity function $v(t)$ with respect to $t$:

$a(t) = \frac{dv}{dt} = \frac{d}{dt}(4t^3 - 18t^2)$

$a(t) = \frac{d}{dt}(4t^3) - \frac{d}{dt}(18t^2)$

$a(t) = (4 \cdot 3 \cdot t^{3-1}) - (18 \cdot 2 \cdot t^{2-1})$

$a(t) = 12t^2 - 36t^1$

$a(t) = 12t^2 - 36t$

Thus, the acceleration of the particle is given by the equation $a(t) = 12t^2 - 36t$.

Comparing with Given Options

Let's compare our calculated acceleration with the given options:

The calculated acceleration equation, $12t^2 - 36t$, matches the equation in Option 4.

Revision Table: Kinematic Quantities

Additional Information on Derivatives in Physics

Derivatives are fundamental tools in physics, especially in kinematics. They allow us to find instantaneous rates of change. For example:

• The first derivative of displacement with respect to time gives instantaneous velocity. This tells us how fast the position is changing at a specific moment.
• The second derivative of displacement (or the first derivative of velocity) with respect to time gives instantaneous acceleration. This tells us how fast the velocity is changing at a specific moment.

In this problem, since the distance $x$ is given as a cubic polynomial in terms of time $t$, the velocity $v$ is a quadratic polynomial in $t$, and the acceleration $a$ is a linear polynomial in $t$. This means the acceleration changes over time, which is characteristic of non-uniform acceleration.