91Ó°ÊÓ

Velocity is a fundamental concept in motion analysis as it describes the rate at which an object changes its position. With differential calculus, we can derive the velocity function from the position function. For a particle moving along a line, this involves differentiating the position function, denoted as \(s(t)\).
To find the velocity \(v(t)\), we differentiate the given position function \(s(t) = (t^2 + 8)e^{-t/3}\) with respect to time \(t\). We apply the product rule because the position function is a product of two functions: \((t^2 + 8)\) and \(e^{-t/3}\). Using the product rule:

• Differentiate \((t^2 + 8)\) to get \(2t\).
• Differentiate \(e^{-t/3}\) to get \(-\frac{1}{3}e^{-t/3}\).

Substitute these derivatives back to get the velocity function: \[v(t) = \left(2t - \frac{t^2 + 8}{3}\right)e^{-t/3}.\]
This expression gives the velocity of the particle at any time \(t\).

Acceleration is another critical concept in differential calculus, measuring how quickly the velocity of an object is changing over time. To find the acceleration function, we take the derivative of the velocity function.
Starting from the velocity function \(v(t) = \left(2t - \frac{t^2 + 8}{3}\right)e^{-t/3}\), we apply the product and chain rules again:

• Differentiating \(\left(2 - \frac{2t}{3}\right)\) gives \(-\frac{2}{3} + \frac{2t}{3}\).
• Differentiating the exponential \(e^{-t/3}\) part gives \(-\frac{1}{3}e^{-t/3}\).

Combining these derivatives, the expression for the acceleration function is: \[a(t) = \left(-\frac{2}{3} + \frac{2t}{3} - \frac{2t^2}{9} - \frac{8}{9}\right)e^{-t/3}.\]
This formula helps us understand how the behavior of the particle's velocity changes over time.

Analyzing particle motion involves understanding various states of an object's movement, such as when it stops, speeds up, or slows down. Let's examine these scenarios step by step for the given function.

• Stopping: The particle stops when its velocity \(v(t)\) is zero. Solve the equation \(v(t) = 0\) to find that the particle stops at \(t = 2\) and \(t = 4\).
• Speeding Up and Slowing Down: The particle's speed is changing based on whether velocity \(v(t)\) and acceleration \(a(t)\) have the same or opposite signs:
  • In interval \((0, 2)\), both \(v(t)\) and \(a(t)\) are negative, indicating the particle is speeding up.
  • In interval \((2, 4)\), the signs can vary, analyzing both velocity and acceleration for specific trends is needed.
  • In interval \(t > 4\), calculate and compare the signs again to determine the motion state.

By analyzing at these intervals, we can deduce the particle's behavior over time.

The product rule is a vital tool in differential calculus, used to differentiate functions that are products of two functions. The rule states that for two differentiable functions \(u(t)\) and \(v(t)\), the derivative of their product is: \[ \frac{d}{dt}[u(t)v(t)] = u(t) \cdot \frac{d}{dt} v(t) + v(t) \cdot \frac{d}{dt} u(t).\]
In our exercise, we applied the product rule to the function \((t^2 + 8)e^{-t/3}\) when deriving both the velocity and acceleration functions.

• We considered \(u(t) = (t^2 + 8)\) and \(v(t) = e^{-t/3}\).
• Apply the rule: differentiate each function independently and then sum the results as per the product rule formula.

This approach allows us to handle complex functions and find their derivatives systematically, which is essential for understanding how particles move over time.