Chapter 4: Problem 2
A point moving on a coordinate line has position function \(s .\) Find the velocity and acceleration at time \(t,\) and describe the motion of the point during the indicated time interval. Illustrate the motion by means of a diagram of the type shown in Figure 4.54 . $$ s(t)=t^{2}+3 t-6 ; \quad[-2,2] $$
Short Answer
Step by step solution
Find the Velocity Function
Find the Acceleration Function
Analyze the Function Over the Interval \\([-2, 2]\\)
Describe the Motion
Construct Motion Diagram
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Position Function
The position function is crucial because it is the foundation for understanding how the object moves. By analyzing the position function, one can deduce patterns or trends in motion, such as changes in direction and speed.
In practical terms, the position of the object at any time \( t \) is found by substituting \( t \) into the equation. This provides a snapshot of the object's location at that particular moment.
Velocity Function
For the position function \( s(t) = t^2 + 3t - 6 \), the velocity function is found by computing the derivative:
- Differentiate to get \( v(t) = s'(t) = 2t + 3 \).
Velocity is essential for understanding motion because it provides information about both the speed and direction of the moving object. Negative velocity indicates movement in one direction (e.g., left), while positive velocity shows motion in the opposite direction (e.g., right).
In this problem, analyzing changes in the velocity helps deduce when and where the object changes direction.
Acceleration Function
Given the velocity function for our exercise, \( v(t) = 2t + 3 \), the derivative provides the acceleration function:
- Differentiate again to get \( a(t) = v'(t) = 2 \).
This constant acceleration means that the velocity increases linearly with time. In this scenario, since the acceleration is positive, it also suggests that the object is continuously speeding up in the direction of positive velocity after a certain point.
Derivative
To find the derivative, apply basic differentiation techniques to each term in the function. For example, for a function \( f(t) = at^n + bt + c \), the derivative is \( f'(t) = ant^{n-1} + b \).
Derivatives find application in many practical contexts, including:
- Determining velocity as the derivative of the position function.
- Finding acceleration by differentiating the velocity function.
Motion Analysis
For the interval \([-2, 2]\) in the given exercise, a comprehensive motion analysis entails:
- Calculating endpoint velocities: \( v(-2) = -1 \), \( v(2) = 7 \).
- Observing the point starts moving left initially then shifts right as time progresses.
Understanding these transitions and how they relate to the calculated values is essential for painting a full picture of motion. Additionally, creating diagrams can provide visual support and clarify changes in speed and direction. Through motion analysis, one captures both the subtle and obvious details of how an object moves over time.