91Ó°ÊÓ

Understanding the motion of an object, such as a mass, involves calculating its position and velocity at particular time points. In this exercise, we're given the position function:
\[ x(t) = 10 \sin(2t) + 15 \cos(2t) + 100 \]
Here, the goal is to determine the position at a specific time, say when \( t = 1 \text{ s} \). Simply substitute \( t = 1 \) into the equation to get:

• Position formula: \( x = 10 \sin(2 \times 1) + 15 \cos(2 \times 1) + 100 \)
• Calculate \( \sin(2) \) and \( \cos(2) \) to find the numerical position.

Velocity is the rate at which the position changes over time. To find the velocity function, take the derivative of the position function with respect to time \( t \). The derivative process gives us:

• For \( 10 \sin(2t) \), derivative is \( 20 \cos(2t) \)
• For \( 15 \cos(2t) \), derivative is \( -30 \sin(2t) \)
• For constant 100, derivative is 0.

Thus, the velocity function becomes:
\[ v(t) = 20 \cos(2t) - 30 \sin(2t) \]
To find the velocity at \( t = 1 \text{ s} \), substitute into the velocity function:
\[ v(1) = 20 \cos(2 \times 1) - 30 \sin(2 \times 1) \]

Differential calculus plays a central role in mechanics, allowing us to describe motion concisely. By differentiating functions related to movement — like position and velocity — we can figure out acceleration, which is the rate of change of velocity.
For this exercise, the velocity function is:
\[ v(t) = 20 \cos(2t) - 30 \sin(2t) \]
Taking the derivative of the velocity function gives the acceleration function \( a(t) \). This process involves applying the rules of differentiation:

• The derivative of \( 20 \cos(2t) \) is \( -40 \sin(2t) \)
• The derivative of \( -30 \sin(2t) \) is \( -60 \cos(2t) \)

Thus, the acceleration function reads:
\[ a(t) = -40 \sin(2t) - 60 \cos(2t) \]
Evaluating this function at \( t = 1 \text{ s} \) gives us the acceleration:
\[ a(1) = -40 \sin(2 \times 1) - 60 \cos(2 \times 1) \]
This use of calculus not only tells us how fast an object is moving but also how its speed is changing.

Finding maximum velocity and acceleration involves analyzing these functions for their peak values. These points are where the object moves fastest or accelerates most strongly.
To locate the points of maximum velocity, set the derivative of the velocity function equal to zero. This means solving:

• Derivative of \( v(t) = 0 \) implies checking \( a(t) = 0 \)
• In this example, \( a(t) = -40 \sin(2t) - 60 \cos(2t) = 0 \)
• Solve to find specific 't' values where maximum velocity occurs.

Similarly, maximum acceleration is considered by finding the maximum magnitude of the acceleration function. Recognizing that sine and cosine functions vary between -1 and 1, the greatest theoretical value occurs when both are at their extremities.
Using the equation \[ \sqrt{(-40)^2 + (-60)^2} \] gives the peak value, as it determines the function's range span based on Pythagorean identity.
This method ensures you can predict both where and how much the velocity and acceleration peak during motion rather than just relying on graphical data.