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In the context of kinematics, a displacement function represents how the position of a particle changes over time. Here, the displacement function given is \(x = \alpha t^2 - \beta t^3\). This formula tells us that the particle's position depends on time \(t\), with coefficients \(\alpha\) and \(\beta\) influencing how the particle moves.

- \(\alpha t^2\) term indicates quadratic growth over time, meaning initially, the particle will move forward increasingly faster.
- \(-\beta t^3\) term suggests a cubic deceleration, which can slow down and eventually reverse the particle's motion.

These terms illustrate the particle starting from the origin (when \(t=0\)), moving forward, and then slowing down as the cubic term outweighs the quadratic term. Analyzing this function allows us to predict important events, like when the particle returns to its starting point or reaches a standstill.

Velocity describes the rate at which the particle's position changes with time. To find it, differentiate the displacement function \(x = \alpha t^2 - \beta t^3\) with respect to \(t\).

Velocity \(v(t)\) is calculated as:
\[ v(t) = \frac{d}{dt}(\alpha t^2 - \beta t^3) = 2\alpha t - 3\beta t^2 \]

This expression tells us that the velocity of the particle varies linearly and quadratically with time.

• The \(2\alpha t\) term indicates that initially, the particle accelerates, adding speed linearly as time progresses.
• The \(-3\beta t^2\) term accounts for a period of deceleration, reducing velocity in a quadratic manner.

Evaluating this differential equation determines moments when the particle may come to rest, like when its velocity equals zero.

Analyzing particle motion involves understanding specific moments that define the motion, such as returning to the starting point, coming to rest, or any special conditions given in the problem.

- **Returning to Start:** For the particle to return to its initial position, solve \(x(t) = 0\):
\[ x(t) = \alpha t^2 - \beta t^3 = 0 \]
Factoring gives:
\[ t^2(\alpha - \beta t) = 0 \]
Solutions include \(t = 0\) or \(t = \alpha/\beta\), indicating the particle returns to the start after \(t = \alpha/\beta\).

- **Coming to Rest:** For the particle to stop, set \(v(t) = 0\):
\[ 2\alpha t - 3\beta t^2 = 0 \]
Factor to find:
\[ t(2\alpha - 3\beta t) = 0 \]
Solutions are \(t = 0\) or \(t = 2\alpha/3\beta\), marking when the particle rests.

Acceleration describes how quickly velocity changes with time. To find this, differentiate the velocity function with respect to time.

Starting from the derived velocity \(v(t) = 2\alpha t - 3\beta t^2\), the acceleration \(a(t)\) is:
\[ a(t) = \frac{d}{dt}(2\alpha t - 3\beta t^2) = 2\alpha - 6\beta t \]

Key insights from this function include:

• Initial acceleration: Evaluate at \(t=0\) to get \(a(0) = 2\alpha\). This shows even at the start, acceleration isn't zero.
• Time of zero acceleration: Set \(a(t) = 0\) to discover times when acceleration vanishes.
  Solve \(2\alpha - 6\beta t = 0\), getting \(t = \alpha/3\beta\). At this instant, no net force acts on the particle, confirming moments of equilibrium.