Problem 15 - Calculus Single Variable

Question

Describes the position of a moving body at time \(t\). Determine whether, at time \(t=4,\) the body is moving forward, backward, or neither. $$ p(t)=-t^{3}+5 t^{2} $$

Step-by-Step Solution

Answer

The body is moving backward at \( t = 4 \).

Step 1: Understand the Position Function

The equation given is \( p(t) = -t^3 + 5t^2 \). This function defines the position \( p \) of a moving body at any given time \( t \). To determine the direction of movement at a specific time, we need to find the velocity of the body at that time.

Step 2: Differentiate Position Function to Find Velocity

Differentiate the position function \( p(t) \) with respect to \( t \) to get the velocity function \( v(t) \). The derivative \( p'(t) \) is: \[ v(t) = p'(t) = \frac{d}{dt}(-t^3 + 5t^2) = -3t^2 + 10t \]. This gives the velocity of the body at any time \( t \).

Step 3: Evaluate Velocity at \( t = 4 \)

Substitute \( t = 4 \) into the velocity function \( v(t) = -3t^2 + 10t \) to find the velocity at that time: \[ v(4) = -3(4)^2 + 10(4) = -48 + 40 = -8 \].

Step 4: Determine Direction of Movement at \( t = 4 \)

The velocity at \( t = 4 \) is \( v(4) = -8 \), which is less than zero. A negative velocity indicates that the body is moving backward at \( t = 4 \).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus used to find the rate at which a quantity changes. When we differentiate a function, we are essentially looking for its derivative. The derivative gives us crucial insights into how the function behaves over time. In the context of motion, differentiating a position function with respect to time gives us the velocity function.

To differentiate the position function in our problem, we use basic differentiation rules. Specifically, if you have a term like \(t^n\), its derivative with respect to \(t\) is \(n \cdot t^{n-1}\). Applying this to each term in the position function \(p(t) = -t^3 + 5t^2\), we find that the derivative is \(v(t) = -3t^2 + 10t\). This new function represents the velocity.

Position Function

The position function describes the location of a moving object at any given time. In our exercise, it's expressed as \(p(t) = -t^3 + 5t^2\). This function helps us understand where the object is in relation to a starting point over time.

Position functions are essential as they serve as the starting point for determining other aspects of motion, such as velocity and acceleration. Typically, the variable \(t\) denotes time, whereas \(p(t)\) represents the position of the object at that particular time. Understanding how these values interact allows us to predict motion behaviors, such as determining whether an object is moving forwards or backwards.

Velocity Function

The velocity function indicates the speed of an object in a particular direction. It is the first derivative of the position function with respect to time and shows how quickly the position changes.

For the position function \(p(t) = -t^3 + 5t^2\), the velocity function came out to be \(v(t) = -3t^2 + 10t\). This indicates that the velocity - or rate of change of position - can vary at different times due to the \(t\) values in the function. To find the velocity at a specific time, you simply substitute the value of \(t\) into the velocity function. Knowing this helps determine how fast and in what direction an object is moving at any given moment.

Direction of Movement

The direction of movement of an object is determined by the sign of the velocity. If the velocity is positive, the object moves forward; if it's negative, it moves backward. When velocity is zero, the object is momentarily at rest.

In the given problem, the velocity function \(v(t) = -3t^2 + 10t\) is used. At \(t = 4\), we evaluate the velocity: \(v(4) = -8\). Since \(-8\) is negative, it clearly indicates the object is moving backward at that time. Understanding the sign of velocity is crucial to understanding not just how fast something is moving, but also in which direction it is moving.