Problem 24 - 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. $$ s=t^{3}-6 t^{2} $$

Step-by-Step Solution

Answer

(a) \( v(t) = 3t^2 - 12t \), \( a(t) = 6t - 12 \). (b) Right: \( t > 4 \). (c) Left: \( 0 < t < 4 \). (d) Negative \( a(t) \): \( 0 \leq t < 2 \). (e) Schematic diagram as described.

Step 1: Find the velocity function v(t)

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

Step 2: Find the acceleration function a(t)

Acceleration is the derivative of the velocity function \( v(t) \) with respect to time. Differentiate \( v(t) = 3t^2 - 12t \) with respect to \( t \).\\[ a(t) = \frac{d}{dt}(3t^2 - 12t) = 6t - 12 \]

Step 3: Determine when the object is moving to the right

The object is moving to the right when its velocity \( v(t) > 0 \). Solve the inequality \( 3t^2 - 12t > 0 \). Factor it as \( 3t(t - 4) > 0 \), which implies \( t > 4 \) or \( t < 0 \). Since time \( t \) is non-negative, the object moves to the right when \( t > 4 \).

Step 4: Determine when the object is moving to the left

The object moves to the left when \( v(t) < 0 \). Solve \( 3t(t - 4) < 0 \). This inequality is satisfied for \( 0 < t < 4 \).

Step 5: Find when acceleration is negative

Acceleration is negative when \( a(t) < 0 \). Solve \( 6t - 12 < 0 \), resulting in \( t < 2 \). So, acceleration is negative for \( 0 \leq t < 2 \).

Step 6: Draw a schematic diagram of the motion

In a schematic diagram:- Mark \( t = 0 \) as the beginning point.- Show the motion to the left from \( 0 < t < 4 \) with negative velocity.- Mark \( t = 4 \) as the point where the object changes direction.- Beyond \( t = 4 \), the object moves to the right as velocity is positive.- From \( 0 \leq t < 2 \), acceleration is negative.

Key Concepts

Velocity Function

The velocity function is a core concept in differential calculus and physics. It tells us how fast an object is moving and in what direction. The velocity function, denoted as \( v(t) \), is derived by calculating the derivative of the position function \( s(t) \) with respect to time \( t \). This differentiation process gives us a function that represents the rate of change of the object's position. For instance, if the position of an object is defined by \( s(t) = t^3 - 6t^2 \), its velocity function becomes \( v(t) = \frac{d}{dt}(t^3 - 6t^2) = 3t^2 - 12t \). A positive value in the velocity function implies movement to the right, a negative value indicates movement to the left, and a zero value means the object is momentarily stationary.

Acceleration Function

Acceleration is a crucial concept that describes how an object's velocity changes over time. It is the derivative of the velocity function \( v(t) \), which itself is already a derivative of the position function. This makes acceleration the second derivative of the position function, \( s(t) \).Computing the acceleration function for the object described by the velocity function \( v(t) = 3t^2 - 12t \) involves differentiating once more: \( a(t) = \frac{d}{dt}(3t^2 - 12t) = 6t - 12 \). Acceleration is positive when \( a(t) > 0 \), indicating the object is speeding up in its current direction, while \( a(t) < 0 \) means the object is slowing down or accelerating in the opposite direction.

Position Function

The position function \( s(t) \) is fundamental in conveying where the object is located along a coordinate line with respect to the origin at any given time \( t \). In this case, the object’s position is given by \( s(t) = t^3 - 6t^2 \). This function allows us to understand the overall path taken by the object.The position function is directly related to both the velocity and acceleration functions since they are its first and second derivatives, respectively. By analyzing \( s(t) \), we can determine when and where on the coordinate line the object changes direction. This happens when the velocity goes from positive to negative or vice versa.

Inequalities in Calculus

Inequalities are an invaluable tool in calculus, especially when analyzing the motion of objects. They help in determining when certain conditions are fulfilled, such as when an object is moving right, left, or when it has negative acceleration.In the context of the exercise, to find when the object moves right, solve the inequality \( 3t^2 - 12t > 0 \), leading to the solution \( t > 4 \). For movement to the left, solve \( 3t^2 - 12t < 0 \), resulting in \( 0 < t < 4 \).Similarly, the inequality for negative acceleration \( a(t) < 0 \) simplifies to \( 6t - 12 < 0 \), indicating \( t < 2 \). These solutions offer insight into the object's behavior over time, showing how inequalities help deduce times of directional and speed changes.