Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

The rate of change of velocity with respect to time is called acceleration. Suppose that the velocity at time \(t\) of a particle is given by \(v(t)=2 t^{2} .\) Find the instantaneous acceleration when \(t=1\) second.

Step-by-Step Solution

Answer

The instantaneous acceleration at \(t=1\) second is 4 \(\text{m/s}^2\).

Step 1: Defining Instantaneous Acceleration

Acceleration is the derivative of velocity with respect to time. Given the function \(v(t) = 2t^2\), the acceleration function \(a(t)\) will be \(a(t) = \frac{dv}{dt}\).

Step 2: Calculating the Derivative

Differentiate the velocity function \(v(t) = 2t^2\) with respect to \(t\): \(\frac{dv}{dt} = \frac{d}{dt}(2t^2) = 4t\).

Step 3: Evaluating at \(t=1\)

Substitute \(t = 1\) into the derivative to find the instantaneous acceleration: \(a(1) = 4 \cdot 1 = 4\).

Key Concepts

Rate of Change of Velocity

When we talk about the rate of change of velocity, we are referring to how quickly an object's speed is increasing or decreasing over time.
Velocity itself is a measure of how fast something is moving in a particular direction, which means it involves both speed and direction.
The rate at which this velocity changes is crucial in understanding acceleration, as acceleration is essentially this rate of change.
For instance, if the velocity of a car increases rapidly, it has a high rate of change, meaning its acceleration is high.
Conversely, if the car's velocity remains constant, the rate of change—and therefore the acceleration—is zero. Here's a simplified way to think about it:

• If the rate of change of velocity is positive, the object is speeding up.
• If the rate is negative, the object is slowing down.
• If the rate is zero, the object's speed is constant.

Understanding this concept helps in solving problems related to motion in physics and calculus, as it provides insight into how forces affect the movement of objects.

Derivative of Velocity

The derivative is a fundamental concept in calculus used to find the rate at which a quantity changes.
When we apply this to velocity, we are looking for the derivative of velocity with respect to time, which gives us acceleration.
This process is mathematical and involves calculus. In our original exercise, we found the derivative of the velocity function, which was given by: \( v(t) = 2t^2 \)By calculating \( \frac{dv}{dt} \), we find that the derivative becomes: \( 4t \).
This function describes the acceleration at any time \( t \).
Here's why differentiation is important:

• It allows us to transition from a function describing velocity to one describing acceleration.
• It offers a tool to predict how velocity will change at any given point in time.
• It gives precise insight into dynamic systems, crucial for advanced problem solving and analysis.

Derivatives are a core part of calculus that empower us to study changes in natural and engineered systems efficiently.

Calculus Problem Solving

Calculus provides powerful tools for solving complex problems involving change. It plays a critical role in understanding motion, growth, and decay in various scientific fields.
By using calculus, we can make sense of how things change and predict future behavior.
The problem-solving process typically involves the following steps:

• Identifying the function or formula e.g., velocity, that represents the initial scenario.
• Finding the derivative of the function to understand the rate of change - in our case, finding acceleration.
• Evaluating the derivative at specific points to gain precise insights - like finding the instantaneous acceleration at \(t = 1\) second.

Applying this to our example, we first identified the velocity function \(v(t) = 2t^2 \).
By differentiating this, we found the function for acceleration.
Evaluating at \(t = 1\) second, we calculated the instantaneous acceleration as \(4\).
Practicing these steps can greatly assist in mastering calculus problem solving.