91Ó°ÊÓ

To delve into velocity calculation, we must first comprehend what velocity means in the context of calculus and kinematics. Velocity refers to the rate at which an object's position changes over time. It is calculated as the first derivative of the position function with respect to time. In this exercise, the position of the object is given by the function \(s = \sin t + \cos t\). By differentiating this with respect to \(t\), we derive the velocity function:

• Velocity function: \(v(t) = \frac{d}{dt}(\sin t + \cos t) = \cos t - \sin t\)

This indicates how quickly the position changes as time progresses. By evaluating this function at a specific point, such as \(t = \frac{\pi}{4}\), we can determine the velocity at that exact moment.
When we substitute \(t = \frac{\pi}{4}\) into the velocity equation \(v(t)\), the results show that:

• \(v\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0\)

Consequently, the velocity at \(t = \frac{\pi}{4}\) is 0 meters/second, meaning that at this point in time, the object is momentarily at rest.

Acceleration is a key element of kinematics. It signifies how the velocity of an object changes with time and is found by taking the derivative of the velocity function.

• Acceleration function: \(a(t) = \frac{d}{dt}(\cos t - \sin t) = -\sin t - \cos t\)

The acceleration function indicates how the velocity’s rate of change differs as time advances. We evaluate the function at \(t = \frac{\pi}{4}\) to find the acceleration for this specific moment.
Substituting \(t = \frac{\pi}{4}\) into the acceleration equation, we find:

• \(a\left(\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2}\)

This tells us that the acceleration at this point in time is \(-\sqrt{2}\) meters/second\(^2\), indicating the velocity is decreasing.
On the other hand, jerk is the rate of change of acceleration. To find jerk, differentiate the acceleration function:

• Jerk function: \(j(t) = \frac{d}{dt}(-\sin t - \cos t) = -\cos t + \sin t\)

By substituting \(t = \frac{\pi}{4}\), we calculate:

• \(j\left(\frac{\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 0\)

Therefore, the jerk at this moment is 0 meters/second\(^3\), signaling that the acceleration is constant around \(t = \frac{\pi}{4}\).

When working with derivatives in the realm of kinematics, it is crucial to understand how to handle trigonometric functions. The position function in this exercise included sine and cosine functions, which have well-known derivatives:

• The derivative of \(\sin t\) is \(\cos t\)
• The derivative of \(\cos t\) is \(-\sin t\)

These derivatives allow us to determine velocity, acceleration, and jerk. They form the backbone of calculations because trigonometric functions often appear in motion-related problems due to their periodic nature.
To better grasp these derivatives:

• Visualize the unit circle and how the sine and cosine values change as you move around it.
• Recognize that the derivative tells us the slope or rate of change of these functions at any point.

Mastering these concepts ensures that you can effectively apply the rules of calculus to real-world scenarios involving periodic movements, like those represented by harmonic oscillations.