Problem 43 - Calculus Single Variable

Question

A position \(p(t)\) is given. Calculate the acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) ). \(p(t)=3 t^{2}+16 t \mathrm{~m}\)

Step-by-Step Solution

Answer

The acceleration is constant at 6 m/s².

Step 1: Understand the Problem

We're given a position function, \(p(t) = 3t^2 + 16t\), which describes the position of an object at time \(t\). Our task is to find the acceleration in \(\mathrm{m/s^2}\). Acceleration is the second derivative of the position function with respect to time \(t\).

Step 2: Find the First Derivative - Velocity

First, we need to find the first derivative of the position function \(p(t)\), which will give us the velocity function \(v(t)\). To do this, differentiate \(p(t) = 3t^2 + 16t\) with respect to \(t\):\[v(t) = \frac{d}{dt}(3t^2 + 16t) = 6t + 16\].

Step 3: Find the Second Derivative - Acceleration

Now, find the second derivative of \(p(t)\), which is the first derivative of the velocity function \(v(t)\). Differentiate \(v(t) = 6t + 16\) with respect to \(t\):\[a(t) = \frac{d}{dt}(6t + 16) = 6\].Thus, the acceleration is \(6\,\mathrm{m/s^2}\).

Step 4: Conclude the Calculation

We determined that the acceleration, which is the second derivative of the position function, is a constant value \(a(t) = 6\) \(\mathrm{m/s^2}\). This means the object has a constant acceleration of \(6\), in the same units as the position was given.

Key Concepts

Position Function

A **position function** describes the location of an object over time. It is often denoted as \(p(t)\) or \(s(t)\), where \(t\) represents time. In this context, our position function is \(p(t) = 3t^2 + 16t\). This function provides an equation that tells us where an object is at any given time. It's like a roadmap indicating the journey of a particle along a line.
To fully understand the position function, think of it as a way to trace how far something has traveled from a starting point at any time \(t\). The formula consists of two parts:

• A quadratic term, \(3t^2\), which contributes to how quickly the position changes over time.
• A linear term, \(16t\), representing a constant velocity component.

This combination allows us to analyze motion in a straightforward way, as we see both uniform and accelerated motion aspects outlined in the equation.

Derivative

The **derivative** is a powerful tool in calculus used to determine the rate at which a function changes. In our exercise, we are particularly interested in the derivatives of the position function to find velocity and acceleration.
When you take the derivative of the position function \(3t^2 + 16t\), it translates mathematical terms into real-world concepts of motion. The first derivative gives us velocity, while the second derivative provides acceleration.
Derivatives can be visually thought of as determining the slope of a curve at any given point. For polynomials like \(p(t)\):

• The derivative of \(3t^2\) is \(6t\), indicating an increasing speed.
• The derivative of \(16t\) is \(16\), a constant reflecting steady motion.

Computing derivatives allows us to predict movement patterns and understand how motion evolves over time.

Velocity

**Velocity** refers to the speed and direction of an object's movement. It is derived as the first derivative of the position function. In our case, the position function \(p(t) = 3t^2 + 16t\) gives us:
\[v(t) = \frac{d}{dt}(3t^2 + 16t) = 6t + 16\]
Here, \(6t\) suggests that the object's speed is increasing as time passes, while the \(16\) indicates a constant rate of change from the initial motion condition.
Velocity not only tells us how fast something is going, but also in what direction it moves. By interpreting the velocity function:

• A positive slope (\(6t\)) combines with a positive constant (\(16\)) signifying an acceleration in the forward direction.
• The direction of motion is consistent since there's no change in sign.

Thus, velocity serves as a crucial link between position and acceleration, showing both speed and trajectory.

Acceleration

**Acceleration** measures how quickly the velocity of an object changes with time. It is the second derivative of the position function. From our position function \(p(t) = 3t^2 + 16t\), we first found the velocity \(v(t) = 6t + 16\). Then, the acceleration is:
\[a(t) = \frac{d}{dt}(6t + 16) = 6\]
This constant \(a(t) = 6\,\mathrm{m/s^2}\) tells us that the object is consistently accelerating at a rate of \(6\,\mathrm{m/s^2}\), meaning each second the velocity increases by 6 meters per second.
Constant acceleration often simplifies real-world calculations, representing steady increases in speed over time without variability. Recognizing acceleration helps interpret how forces impact motion, explaining how quickly dynamic situations can evolve.