91Ó°ÊÓ

Velocity is a key concept in understanding particle motion. It tells us how fast a particle is moving and in what direction. To find velocity, we differentiate the position function with respect to time. In our exercise, the position of the particle is given by the function \[ x = (t-2)^2(t-5) \].To calculate the velocity, we use the derivative of this function. This involves applying the product rule of differentiation, which handles the multiplication of two functions. For our function, it looks like this: \[ v(t) = \frac{d}{dt}[(t-2)^2(t-5)]. \]After applying the differentiation rules, we simplify to find the velocity function, \[ v(t) = (t-2)(3t-12). \]Next, to find out exactly what the velocity is at a specific time, say \( t = 2 \), we substitute this value into our velocity function:\[ v(2) = (2-2)(3 \cdot 2 - 12) = 0. \]So, the velocity at \( t = 2 \) is 0, indicating the particle is momentarily at rest.

Acceleration tells us how the velocity of a particle changes over time. Like velocity, the acceleration is calculated by differentiating the velocity function with respect to time. From our previous calculation, we have the velocity function: \[ v(t) = (t-2)(3t-12). \]To find the acceleration, we differentiate this function: \[ a(t) = \frac{d}{dt}[(t-2)(3t-12)]. \]Applying the product rule, the acceleration function simplifies to \[ a(t) = 6t - 18. \]This equation describes how the acceleration varies with time. To check if the acceleration is negative at certain times, substitute a value into \( a(t) \). For example, for the time \( t = 2 \): \[ a(2) = 6 \cdot 2 - 18 = -6. \]A negative result indicates the particle is slowing down. Indeed, for \( t < 3 \), acceleration is negative, meaning the particle decelerates during this interval.

Differentiation is a fundamental tool in physics, especially in the study of motion. Here, it helps us transition from position to velocity and from velocity to acceleration. The cornerstone idea is that differentiation allows us to understand how a quantity changes instantaneously at any given point in time.
When studying motion, the first derivative of the position function with respect to time provides the velocity. This step requires one to apply differentiation rules, such as the product rule, which is useful when the position is defined by a product of functions. The velocity, as shown, was derived to be:\[ v(t) = (t-2)(3t-12). \]
Following this, the second derivative - or the derivative of the velocity - gives us the acceleration. Again, complex functions require careful consideration of differentiation rules to ensure accurate results. In our example, this provided:\[ a(t) = 6t - 18. \]
Through differentiation, we can gain insights into not just where a particle is, but also how fast it moves and how its speed changes over time - essential aspects in describing motion fully.