[Technology Exercise] Exercises 31–34 give the position function s = f(t) of an object moving along the s-axis as a function of time t. Graph f together with the velocity function v(t) = ds/dt = f'(t) and the acceleration function a(t) = d²s/dt² = f''(t). Comment on the object’s behavior in relation to the signs and values of v and a. Include in your commentary such topics as the following:
c. When does it change direction?
s = t² - 3t + 2, 0 ≤ t ≤ 5
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Step 1: Identify the position function s(t) = t² - 3t + 2. This function describes the position of the object along the s-axis over time.
Step 2: Find the velocity function v(t) by differentiating the position function s(t) with respect to time t. This gives v(t) = ds/dt = f'(t) = 2t - 3.
Step 3: Determine the acceleration function a(t) by differentiating the velocity function v(t) with respect to time t. This results in a(t) = dv/dt = f''(t) = 2.
Step 4: Analyze when the object changes direction by finding when the velocity function v(t) = 0. Solve the equation 2t - 3 = 0 to find the critical points.
Step 5: Comment on the object's behavior: The object changes direction at the time when v(t) = 0. Since the acceleration a(t) = 2 is constant and positive, it indicates that the object is accelerating in the positive direction throughout the interval 0 ≤ t ≤ 5.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Position, Velocity, and Acceleration
In calculus, the position function s(t) describes the location of an object over time. The velocity function v(t) is the first derivative of the position function, representing the rate of change of position, or speed and direction. Acceleration a(t) is the second derivative of the position function, indicating the rate of change of velocity. Understanding these relationships is crucial for analyzing motion.
The derivative of a function at a point gives the slope of the tangent line to the function at that point, representing the rate of change. For motion, the first derivative of the position function gives velocity, and the second derivative gives acceleration. Analyzing these derivatives helps determine when an object speeds up, slows down, or changes direction.
The signs of velocity and acceleration provide insights into an object's motion. A positive velocity indicates forward motion, while a negative velocity indicates backward motion. When velocity changes sign, the object changes direction. Acceleration's sign shows whether the object is speeding up or slowing down; positive acceleration increases velocity, while negative acceleration decreases it.