Problem 37 - Calculus Single Variable

Question

A velocity \(v(t)\) is given. Calculate the acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) ). \(v(t)=24 \mathrm{~m} / \mathrm{s}\)

Step-by-Step Solution

Answer

The acceleration is 0 m/s² since the velocity is constant.

Step 1: Understand the Relationship

To find acceleration from velocity, we utilize the relationship that acceleration is the derivative of velocity with respect to time. Mathematically, this is expressed as \( a(t) = \frac{dv}{dt} \), where \( a(t) \) is the acceleration.

Step 2: Determine the Derivative

The function given is \( v(t) = 24 \mathrm{~m/s} \). This is a constant function, meaning it does not change with time.

Step 3: Compute the Derivative

Since \( v(t) \) is constant, its derivative with respect to time is 0. Therefore, we have \( \frac{dv}{dt} = 0 \).

Step 4: State the Acceleration

The acceleration \( a(t) \) is the derivative of \( v(t) \), which we calculated as 0. Thus, \( a(t) = 0 \mathrm{~m/s^2} \).

Key Concepts

Velocity

Velocity is a crucial concept in calculus and physics, as it describes the speed of an object in a specified direction. Imagine a car traveling down a straight road—the velocity tells you not only how fast it's going but also in which direction. Mathematically, velocity is a function of time, often denoted as \( v(t) \), where \( t \) represents time. This means, at different times, the object could have different velocities.

• The velocity function allows us to understand motion by indicating changes in speed and direction over time.
• Units for velocity are typically meters per second (m/s).

In our exercise, the velocity \( v(t) = 24 \, \mathrm{m/s} \) indicates a constant velocity since it does not depend on time. This implies that the object's speed and direction are unchanging.

Acceleration

Acceleration describes how the velocity of an object changes with time. It's like noticing if a car speeds up, slows down, or makes a turn. Acceleration is the rate at which velocity changes and is measured in meters per second squared (m/s²). In calculus, acceleration is found by taking the derivative of the velocity function with respect to time.

• Positive acceleration means the object is speeding up.
• Negative acceleration (sometimes called deceleration) means the object is slowing down.
• Zero acceleration implies constant velocity, meaning no change in speed or direction.

In our example, since the velocity is constant, the calculated acceleration is \( 0 \, \mathrm{m/s^2} \). This indicates there is no change in motion.

Derivative

Derivatives are a fundamental tool in calculus that help us determine how a function changes as its input changes. For motion-related functions like velocity and acceleration, the derivative gives insight into how one aspect of motion changes over time.

• The derivative of position (distance function) with respect to time gives the velocity.
• The derivative of the velocity function gives the acceleration.

Mathematically, if you have a function \( v(t) \), the derivative \( \frac{dv}{dt} \) tells us the acceleration. In the case of a constant velocity \( v(t) = 24 \, \mathrm{m/s} \), the derivative is \( 0 \) because constants do not change, meaning no acceleration.

Constant Function

A constant function is one that does not change regardless of the input. Think of it as a horizontal line when graphed on a coordinate plane. In the context of our problem, \( v(t) = 24 \, \mathrm{m/s} \) is a constant function because the velocity remains the same across all times \( t \).

• In calculus, the derivative of a constant function is always zero because there is no change to measure.
• This feature simplifies calculations significantly, especially when analyzing motion.

Understanding when a function is constant can help quickly determine related quantities like acceleration, as we observed in this exercise.