https://file.aiquickdraw.com/mathbot/file/1727203233380b6w5lmw6.jpg

From the problem description, we have different phases of driving described:

1. **Acceleration (0 ≤ t ≤ 20 seconds)**: The driver gradually increases the force on the pedal, implying an increasing acceleration. This means the second derivative of the position function \( s''(t) \) (acceleration) is positive.
2. **Deceleration (20 < t ≤ 40 seconds)**: The driver begins to gradually lift the foot off the pedal at \( t = 20 \) seconds and continues decreasing the force, culminating in stopping at \( t = 40 \) seconds. This phase implies a decreasing acceleration, or deceleration, where \( s''(t) \) becomes negative.

To find when the second derivative switches from positive to negative, we need to identify the exact point \( t = 20 \) seconds. At this point, the transition from accelerating to braking occurs, which signifies the change in the sign of \( s''(t) \) from positive to negative. Therefore, the answer to when the second derivative switches sign is at \( t = 20 \) seconds.

Would you like more details on how the acceleration and deceleration processes relate to the second derivative or any other mathematical concepts? Here are some follow-up questions you might consider:

1. How can we calculate the actual value of acceleration at a specific time \( t \) if given the force applied?
2. What is the significance of the second derivative being zero at any point?
3. What could be the graphical representation of \( s(t) \), \( s'(t) \), and \( s''(t) \) for this scenario?
4. How does the change in acceleration affect the overall journey time and distance?
5. What are some real-world implications of understanding acceleration and deceleration in vehicle dynamics?

**Tip**: Understanding the second derivative in terms of physical movements can significantly aid in predicting system behaviors in physics and engineering applications.

Suppose you are driving a car. Let s(t) denote your position in feet at time t (and assume s(t) is differentiable). Consider an interval of time [0, 40] (in seconds). At t = 0, you put your foot on the pedal softly and keep the same amount of force on the pedal for 10 seconds. At the 10-second mark, you begin increasing the force of your foot gradually until t = 20, when you see a red light in the distance, so you gradually lift your foot from the pedal. At t = 30, you begin braking, and finally stop at the red light at t = 40. At what time t does the sign of the second derivative switch from positive to negative or from negative to positive?

Explore the calculus-based problem involving a car's motion. Understand how the second derivative of the position function s(t) switches from positive to negative, indicating the transition from acceleration to deceleration. Suitable for college-level calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation