Problem 150 - Calculus Volume 1

Question

For the following exercises, the given functions represent the position of a particle traveling along a horizontal line. a. Find the velocity and acceleration functions. b. Determine the time intervals when the object is slowing down or speeding up. $$s(t)=2 t^{3}-3 t^{2}-12 t+8$$

Step-by-Step Solution

Answer

The velocity is \(6t^2 - 6t - 12\) and acceleration is \(12t - 6\). Object speeds up on \((-\infty, -1)\) and \((2, \infty)\); slows down on \((-1, 2)\).

Step 1: Find the Velocity Function

The velocity function is the derivative of the position function. To find the velocity function, differentiate the position function \( s(t) = 2t^3 - 3t^2 - 12t + 8 \) with respect to \( t \). \[v(t) = \frac{d}{dt}(2t^3 - 3t^2 - 12t + 8) = 6t^2 - 6t - 12\]The velocity function is \( v(t) = 6t^2 - 6t - 12 \).

Step 2: Find the Acceleration Function

The acceleration function is the derivative of the velocity function. Differentiate the velocity function \( v(t) = 6t^2 - 6t - 12 \) with respect to \( t \). \[a(t) = \frac{d}{dt}(6t^2 - 6t - 12) = 12t - 6\]The acceleration function is \( a(t) = 12t - 6 \).

Step 3: Determine When the Object is Speeding Up or Slowing Down

An object is speeding up when the signs of the velocity and acceleration are the same, and slowing down when the signs are different.First, find the critical points for velocity turning zero by setting \( v(t) = 0 \):\[6t^2 - 6t - 12 = 0\]Divide the entire equation by 6:\[t^2 - t - 2 = 0\]Factor the quadratic:\[(t-2)(t+1)=0\]This gives \( t = 2 \) and \( t = -1 \).Next, analyze the intervals formed by these points and the acceleration:- For \( t < -1 \): - \( v(t) < 0 \) and \( a(t) < 0 \) (both negative), object is speeding up.- For \( -1 < t < 2 \): - \( v(t) > 0 \) and \( a(t) < 0 \) (opposite signs), object is slowing down.- For \( t > 2 \): - \( v(t) > 0 \) and \( a(t) > 0 \) (both positive), object is speeding up.Thus, the object speeds up on intervals \( (-\infty, -1) \) and \( (2, \infty) \) and slows down on \( (-1, 2) \).

Key Concepts

Derivatives

In calculus, derivatives represent the rate at which a function is changing at any given point. It's a fundamental tool for understanding how functions behave. In our exercise, we were given the position function of a particle moving along a line. To find how fast and in what direction the particle is moving, we calculate its derivative—known as the velocity function. For the position function \( s(t) = 2t^3 - 3t^2 - 12t + 8 \), the velocity function \( v(t) \) is derived by differentiating \( s(t) \) with respect to time \( t \). This involves using the power rule for differentiation. The power rule states that if you have a term like \( at^n \), its derivative is \( nat^{n-1} \). Applying this to our position function, we get \( v(t) = 6t^2 - 6t - 12 \).

Understanding derivatives allows us to interpret the motion of objects. Just as we derive the velocity from position, we can further find acceleration by taking the derivative of the velocity function.

Velocity and Acceleration

Velocity tells us how fast an object is moving and in which direction, while acceleration provides the rate of change of velocity. When differentiating the velocity function \( v(t) = 6t^2 - 6t - 12 \), we obtain the acceleration function \( a(t) = 12t - 6 \).

Acceleration can indicate whether an object is speeding up or slowing down:

• If the signs of velocity and acceleration are the same, the object speeds up. This occurs because they complement each other, pushing the object forward in one consistent direction.
• If the signs are opposite, the object slows down, as the acceleration is acting against the velocity.

By analyzing both functions together, we can form a detailed understanding of the object's motion over time.

Dynamic Motion Analysis

Dynamic motion analysis in calculus involves studying how the velocity and acceleration of an object change over time. This analysis helps us determine intervals during which an object speeds up or slows down. To achieve this, we solve for the critical points where the velocity is zero by setting \( v(t) = 6t^2 - 6t - 12 = 0 \). Factoring this equation provides the critical points \( t = -1 \) and \( t = 2 \).

We then analyze the intervals created by these critical points. For instance:

• In the interval \( (-\infty, -1) \), both velocity and acceleration are negative, which means the object is speeding up.
• Between \( (-1, 2) \), the velocity is positive but the acceleration is negative, resulting in the object slowing down.
• For \( (2, \infty) \), both velocity and acceleration are positive, indicating the object speeds up again.

Dynamic motion analysis not only helps us determine when changes in speed occur but also provides insights into how different forces might be affecting the object's trajectory.