Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 23-28, an object is moving along a horizontal coordinate line according to the formula \(s=f(t)\), where \(s\), the directed distance from the origin, is in feet and \(t\) is in seconds. In each case, answer the following questions (see Examples 2 and 3). (a) What are \(v(t)\) and \(a(t)\), the velocity and acceleration, at time \(t\) ? (b) When is the object moving to the right? (c) When is it moving to the left? (d) When is its acceleration negative? (e) Draw a schematic diagram that shows the motion of the object. $$ s=2 t^{3}-6 t+5 $$

Step-by-Step Solution

Answer

(a) \( v(t) = 6t^2 - 6 \), \( a(t) = 12t \); (b) right for \( t > 1 \) and \( t < -1 \); (c) left for \( -1 < t < 1 \); (d) negative for \( t < 0 \).

Step 1: Differentiate to Find Velocity

To find the velocity function, differentiate the position function \( s(t) = 2t^3 - 6t + 5 \) with respect to \( t \). The derivative of the position function \( s(t) \) with respect to time \( t \) will give us the velocity function \( v(t) \).\[ v(t) = \frac{d}{dt}(2t^3 - 6t + 5) = 6t^2 - 6 \]

Step 2: Differentiate to Find Acceleration

To find the acceleration function, differentiate the velocity function \( v(t) = 6t^2 - 6 \) with respect to \( t \). The derivative of the velocity function will give us the acceleration function \( a(t) \).\[ a(t) = \frac{d}{dt}(6t^2 - 6) = 12t \]

Step 3: Determine When the Object is Moving Right

The object is moving to the right when the velocity \( v(t) > 0 \). Set the velocity function greater than zero and solve for \( t \).\[ 6t^2 - 6 > 0 \]\[ 6(t^2 - 1) > 0 \]\[ t^2 > 1 \]\[ t > 1 \text{ or } t < -1 \]

Step 4: Determine When the Object is Moving Left

The object moves to the left when the velocity \( v(t) < 0 \). Set the velocity function less than zero and solve for \( t \).\[ 6t^2 - 6 < 0 \]\[ 6(t^2 - 1) < 0 \]\[ t^2 < 1 \]\[ -1 < t < 1 \]

Step 5: Determine When the Acceleration is Negative

The acceleration is negative when \( a(t) < 0 \). Set the acceleration function less than zero and solve for \( t \).\[ 12t < 0 \]\[ t < 0 \]

Step 6: Draw Schematic Diagram

Create a diagram that shows the behavior of the object. This includes marking the regions where the object moves left, right, and where the acceleration is negative. Since descriptions rely on visual representations, a schematic diagram highlights the critical points of the object's motion.

Key Concepts

Derivatives

In calculus, derivatives help us understand how a function changes at any given point. For motion, this often translates into how the position of an object changes over time.

• The derivative of a function gives us the rate of change of that function.
• In motion problems, taking the derivative of a position function with respect to time gives us the velocity function.

For the given problem, we start with a position function: \[ s(t) = 2t^3 - 6t + 5 \]By taking the derivative of this function, we find how fast the position \( s(t) \) is changing with time. Thus, the velocity \( v(t) \) is\[ v(t) = 6t^2 - 6 \]Derivatives give us a powerful tool to predict motion and understand the dynamics of moving objects.

Velocity

Velocity describes the speed and direction of a moving object.

• It is the first derivative of the position function with respect to time.
• Velocity is a vector, meaning it has both magnitude and direction.

From our example, the velocity function is derived as\[ v(t) = 6t^2 - 6 \]This tells us how fast and in which direction the object is moving at any time \( t \). When \( v(t) > 0 \), the object is moving to the right, and when \( v(t) < 0 \), it is moving to the left.

• For \( t > 1 \) or \( t < -1 \), the velocity is positive, indicating movement to the right.
• For \( -1 < t < 1 \), the velocity is negative, indicating movement to the left.

Understanding velocity helps us figure out not just speed, but the direction of an object's movement over time.

Acceleration

Acceleration is the rate of change of velocity. It's the derivative of the velocity function with respect to time. In other words, it tells us how quickly an object is speeding up or slowing down.

• Acceleration is also a vector quantity.
• If velocity is changing, acceleration is present.

For our problem, by differentiating the velocity function \( v(t) = 6t^2 - 6 \), we get\[ a(t) = 12t \]This function helps determine the object's acceleration at any time \( t \). We note that:

• When \( a(t) < 0 \), the acceleration is negative, indicating the object is slowing down. This happens for \( t < 0 \).

Understanding acceleration is crucial to predicting changes in the motion and force applied on moving objects.

Motion analysis

Motion analysis involves understanding how an object moves across different dimensions over time.

• This includes analyzing velocity and acceleration over a specific time interval.
• By understanding its motion, we can make predictions about its future behavior.

By solving the equations from our textbook problem, we can conclude:- The object moves right for \( t > 1 \) and \( t < -1 \).- It moves left for \( -1 < t < 1 \).- Acceleration is negative for \( t < 0 \).These insights allow us to create a schematic diagram of motion. Visualizing these patterns helps understand complex motion dynamics easier, setting a foundation for more advanced physics and engineering concepts. By mapping out when the object moves left or right, and when it accelerates or decelerates, you're laying the groundwork for accurate motion prediction.