Problem 39 - Calculus Single Variable

Question

A velocity \(v(t)\) is given. Calculate the acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) ). \(v(t)=-16 t+5 \mathrm{~m} / \mathrm{s}\)

Step-by-Step Solution

Answer

The acceleration is constant and equals \(-16 \mathrm{~m/s}^2\).

Step 1: Understanding the relationship between velocity and acceleration

Acceleration is the derivative of velocity with respect to time. To find acceleration, we need to differentiate the given velocity function \(v(t)\) with respect to \(t\).

Step 2: Differentiate the velocity function

The velocity function given is \(v(t) = -16t + 5\). Differentiate this with respect to \(t\).\[\frac{dv}{dt} = \frac{d}{dt}(-16t + 5)\]Differentiating term-by-term gives:\[-16\cdot \frac{d}{dt}t + 0 = -16\times 1 = -16\]

Step 3: Interpret the result

The derivative \(-16\) indicates that the acceleration is constant and does not depend on time. The units of this acceleration are in \(\mathrm{m/s}^2\).

Key Concepts

Velocity Function

In the context of motion, the velocity function describes how an object's velocity changes over time. Velocity itself is a vector quantity, meaning it has both magnitude and direction. When we have a velocity function, such as \(v(t) = -16t + 5\), it represents how fast something is moving and in which direction at any given time \(t\).

• The term \(-16t\) indicates the velocity changes as time progresses.
• The constant value \(+5\) represents the initial velocity at time \(t=0\).

Understanding the velocity function is essential for analyzing the motion, as it provides information about how the speed and direction of an object change. In this exercise, this function is linear, implying the velocity changes at a constant rate, leading us to the next concept: acceleration.

Derivative

The derivative is a fundamental concept in calculus representing the rate at which a function changes. When applied to physical concepts, it allows us to understand changes over time, like velocity and acceleration. In our scenario, we differentiate the velocity function to find acceleration.

• Derivatives help identify slopes and rates of change.
• For the function \(v(t) = -16t + 5\), the derivative tells us how velocity itself changes with time, i.e., the acceleration.

Using differentiation, we find that the derivative of \(-16t\) with respect to time is \(-16\). This computation indicates the constant slope of the velocity function, reflecting constant acceleration.

Constant Acceleration

Constant acceleration means that an object's change in velocity remains steady over time. In our exercise, differentiating the velocity function \(v(t) = -16t + 5\) yields \(-16\). This result signifies that the acceleration is constant at \(-16 \ \mathrm{m/s}^2\).

• This constancy implies that the object's velocity decreases at a steady rate each second.
• Constant acceleration often simplifies analyzing motion, as the calculations remain consistent over time.

With a constant acceleration, prediction of future velocity or position becomes straightforward using linear equations derived from initial values.

Differentiation

Differentiation is the process of finding a derivative, and it provides crucial insights into how things change over time. By differentiating the velocity function, we determine the acceleration.

• The steps involve computing derivative term-by-term.
• For each term in \(v(t) = -16t + 5\), differentiate separately: the derivative of \(-16t\) is \(-16\), and the derivative of the constant 5 is 0.

The use of differentiation provides a systematic way to transition from a velocity function to an expression for acceleration, helping unravel the physics of motion by quantifying how swiftly velocity itself changes.