Moment When The Car Spesi Car At That Moment. Question 4. By Using Derivative, Determine The Intervals
Understanding the behavior of a moving vehicle, especially when analyzing its acceleration, deceleration, or constant speed phases, is crucial in physics and calculus. In particular, the question of when a car reaches specific speeds or changes its velocity significantly can be precisely answered using the tools of calculus—most importantly, derivatives. This article explores how to determine the intervals during which a car is speeding up, slowing down, or moving at a constant velocity by applying derivative concepts.
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Introduction to Derivatives in Motion Analysis
The velocity of a car at any given moment is represented mathematically by the derivative of its position function with respect to time. If we denote the position of the car at time \( t \) as \( s(t) \), then:
\[
v(t) = s'(t) = \frac{ds}{dt}
\]
Similarly, the acceleration is the derivative of velocity with respect to time:
\[
a(t) = v'(t) = s''(t)
\]
Using these derivatives, we can analyze the behavior of the car at different time intervals, understanding where it accelerates, decelerates, or maintains a constant speed.
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Understanding the Key Concepts
Critical Points and Their Significance
A critical point occurs at a time \( t \) where the derivative \( v(t) \) is zero or undefined:
- If \( v(t) = 0 \), the car is momentarily at rest.
- If \( v(t) \) changes sign around \( t \), the car switches from moving forward to backward or vice versa.
Identifying these points helps us understand transition moments in the car's motion.
Intervals of Increase and Decrease
- Intervals where \( v(t) > 0 \): The car is moving forward and increasing its position.
- Intervals where \( v(t) < 0 \): The car is moving backward or decreasing its position.
- Intervals where \( v(t) = 0 \): The car is stationary or at a turning point.
Concavity and Inflection Points
While primarily used to analyze the shape of the position graph, the second derivative \( a(t) \) indicates whether the velocity is increasing or decreasing.
- Positive \( a(t) \): The velocity is increasing; the car is speeding up.
- Negative \( a(t) \): The velocity is decreasing; the car is slowing down.
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Step-by-Step Approach to Determine Intervals Using Derivatives
To analyze the intervals where the car speeds up or slows down, follow these systematic steps:
1. Obtain the Position Function \( s(t) \)
The problem typically provides a function \( s(t) \), representing the position of the car at time \( t \). For example:
\[
s(t) = at^3 + bt^2 + ct + d
\]
or any other differentiable function.
2. Find the Velocity Function \( v(t) = s'(t) \)
Differentiate \( s(t) \) with respect to \( t \):
\[
v(t) = \frac{ds}{dt}
\]
This gives the rate of change of position, or velocity at any moment.
3. Determine Critical Points of \( v(t) \)
Solve \( v(t) = 0 \) to find potential points where the car changes behavior:
\[
v(t) = 0
\]
Use algebraic methods or numerical techniques depending on the function's complexity.
4. Evaluate Sign of \( v(t) \) Around Critical Points
Test points in each interval between critical points to determine whether \( v(t) \) is positive or negative. This indicates the direction of motion.
5. Find the Acceleration Function \( a(t) = v'(t) \)
Differentiate \( v(t) \):
\[
a(t) = \frac{dv}{dt} = s''(t)
\]
This helps determine where the car is speeding up or slowing down.
6. Analyze the Sign of \( a(t) \) in Each Interval
- If \( a(t) > 0 \) and \( v(t) > 0 \), the car is speeding up.
- If \( a(t) < 0 \) and \( v(t) > 0 \), the car is slowing down.
- Similar logic applies when the velocity is negative.
Applying Derivatives: An Example Problem
Suppose the position function of a car is given by:
\[
s(t) = 2t^3 - 9t^2 + 12t
\]
where \( s(t) \) is in meters, and \( t \) in seconds.
Step 1: Find \( v(t) \)
\[
v(t) = s'(t) = 6t^2 - 18t + 12
\]
Step 2: Determine critical points for \( v(t) = 0 \)
\[
6t^2 - 18t + 12 = 0
\]
Divide through by 6:
\[
t^2 - 3t + 2 = 0
\]
Factor:
\[
(t - 1)(t - 2) = 0
\]
Critical points at:
\[
t = 1, \quad t = 2
\]
Step 3: Sign analysis of \( v(t) \)
- For \( t < 1 \), pick \( t=0.5 \):
\[
v(0.5) = 6(0.5)^2 - 18(0.5) + 12 = 6(0.25) - 9 + 12 = 1.5 - 9 + 12 = 4.5 > 0
\]
- For \( 1 < t < 2 \), pick \( t=1.5 \):
\[
v(1.5) = 6(2.25) - 18(1.5) + 12 = 13.5 - 27 + 12 = -1.5 < 0
\]
- For \( t > 2 \), pick \( t=3 \):
\[
v(3) = 6(9) - 18(3) + 12 = 54 - 54 + 12 = 12 > 0
\]
Interpretation:
- The car moves forward (positive velocity) during \( t \in [0,1) \).
- It moves backward (negative velocity) during \( t \in (1,2) \).
- It moves forward again during \( t > 2 \).
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Analyzing Acceleration to Determine Speeding Up and Slowing Down
Step 4: Find \( a(t) = v'(t) \)
\[
a(t) = \frac{d}{dt}(6t^2 - 18t + 12) = 12t - 18
\]
Step 5: Find when \( a(t) = 0 \)
\[
12t - 18 = 0 \implies t = 1.5
\]
Step 6: Sign analysis of \( a(t) \)
- For \( t < 1.5 \), say \( t=1 \):
\[
a(1) = 12(1) - 18 = -6 < 0
\]
- For \( t > 1.5 \), say \( t=2 \):
\[
a(2) = 24 - 18 = 6 > 0
\]
Interpretation:
- When \( a(t) < 0 \), the velocity is decreasing; the car is slowing down.
- When \( a(t) > 0 \), the velocity is increasing; the car is speeding up.
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Summarizing the Intervals of Speeding Up and Slowing Down
Based on the above analysis:
| Interval | Velocity \( v(t) \) | Acceleration \( a(t) \) | Motion Status | Speeding Up or Slowing Down |
|------------|---------------------|-------------------------|---------------------------|----------------------------------------|
| \( t \in [0,1) \) | Positive | Negative (\( a(t)<0 \)) | Moving forward | Slowing down (velocity decreasing) |
| \( t \in (1,1.5) \) | Negative | Negative | Moving backward | Slowing down |
| \( t \in (1.5,2) \) | Negative | Positive | Moving backward | Speeding up (velocity decreasing in magnitude) |
| \( t > 2 \) | Positive | Positive | Moving forward | Speeding up |
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Practical Implications and Real-World Applications
Understanding the intervals when a car accelerates or decelerates has multiple practical applications:
- Safety Analysis: Detecting when a vehicle is speeding up or slowing down helps in accident prevention.
- Performance Optimization: Car engineers analyze these intervals to improve
