Problem 33 - Single Variable Calculus

Question

Acceleration of a Car A car starting from rest and traveling in a straight line attains a velocity of $$v(t)=\frac{110 t}{2 t+5}$$ feet per second after \(t\) sec. Find the initial acceleration of the car and its acceleration \(10 \mathrm{sec}\) after starting from rest.

Step-by-Step Solution

Answer

The initial acceleration of the car is \(\frac{220}{25}\) feet per second squared, and the acceleration of the car 10 seconds after starting from rest is \(\frac{-4180}{625}\) feet per second squared.

Step 1: Find the acceleration function by differentiating the velocity function

Recall that acceleration is the derivative of velocity with respect to time. We have the velocity function: \(v(t) = \frac{110t}{2t + 5}\) Now, we need to find its derivative with respect to time, let's call it \(a(t)\). To differentiate this function, we'll use the quotient rule, which states: \(\frac{d}{dt} \left(\frac{u}{v}\right) = \frac{vu' - uv'}{v^2}\) In our case, \(u = 110t\) and \(v = 2t + 5\). Then, the derivatives with respect to time are: \(u' = \frac{d}{dt}(110t) = 110\) \(v' = \frac{d}{dt}(2t + 5) = 2\) Now we can apply the quotient rule to find the derivative of the velocity function: \(a(t) = \frac{(2t + 5)(110) - (110t)(2)}{(2t + 5)^2}\) Simplify the equation: \(a(t) = \frac{220 - 440t}{(2t + 5)^2}\)

Step 2: Find the initial acceleration

To find the initial acceleration, we'll evaluate \(a(t)\) at \(t = 0\): \(a(0) = \frac{220 - 440(0)}{(2(0) + 5)^2} = \frac{220}{25}\) Thus, the initial acceleration of the car is \(\frac{220}{25}\) feet per second squared.

Step 3: Find the acceleration at 10 seconds

To find the acceleration after 10 seconds, we'll evaluate \(a(t)\) at \(t = 10\): \(a(10) = \frac{220 - 440(10)}{(2(10) + 5)^2} = \frac{-4180}{625}\) Thus, the acceleration of the car 10 seconds after starting from rest is \(\frac{-4180}{625}\) feet per second squared.

Key Concepts

Derivative of Velocity

The derivative of velocity with respect to time is a fundamental concept in calculus and physics. It is the rate at which velocity changes over time, and in the realm of mechanics, it is termed as acceleration. When we have a velocity function, such as in our car example,



\(v(t) = \frac{110t}{2t + 5}\), we look for its derivative to understand how quickly the car is speeding up or slowing down as time passes. The action of taking the derivative is also called differentiation, which we'll explore more in-depth further in this article. Differentiating correctly requires the knowledge of rules and formulas specific to the function at hand. In this case, we'll be using the quotient rule to find the initial acceleration. Understanding the derivative of velocity is crucial in determining how an object's motion changes over time.

Acceleration Function

An acceleration function represents the acceleration of an object as a function of time. It is derived from the velocity function of the object, as the velocity is a measure of how fast the object is moving and acceleration is how fast that speed itself is changing. In the given problem, we found the acceleration function

\(a(t)\) by differentiating the velocity function \(v(t)\). The acceleration function gives us a formula to calculate the rate of change of velocity at any given time. For initial acceleration, when \(t = 0\), it shows how quickly the car starts moving from rest. Further, by evaluating the acceleration function at any specific time, such as \(t = 10\) seconds in our example, we can determine the car's acceleration at that moment.

Quotient Rule

The quotient rule is a method used in calculus to differentiate functions that are ratios of two differentiable functions. It is expressed as \(\frac{d}{dt} \left(\frac{u}{v}\right) = \frac{vu' - uv'}{v^2}\) where \(u\) and \(v\) are functions of \(t\), and \(u'\) and \(v'\) are their respective derivatives. When applying the quotient rule to the velocity function in our car example, we identify \(u = 110t\) and \(v = 2t + 5\), with their derivatives being \(u' = 110\) and \(v' = 2\). The quotient rule enables us to find the acceleration function, which is the velocity function's derivative. It is particularly useful when dealing with rates of change in ratios, as it's often the case in physics and engineering problems.

Differentiation

Differentiation is the process of finding a derivative, which is a measure of how a function changes as its input changes. It's a fundamental tool in calculus that allows us to calculate the rate of change of one quantity with respect to another. Differentiation can be used to determine the slope of a curve at any point, and when it comes to motion, it helps find acceleration or deceleration at any given time. The process involves rules like the power rule, product rule, chain rule, and as seen in our car example, the quotient rule. Each of these rules applies to different kinds of functions to ensure accurate differentiation. Mastering differentiation is essential for solving a wide range of physical and mathematical problems.