Problem 220 - Calculus Volume 1

Question

[T] A race car is accelerating at a velocity given by $$ v(t)=\frac{25}{4} t+54 $$ where \(v\) is the velocity (in feet per second) at time \(t\) a. Find the velocity of the car at 10 \(\mathrm{sec}\) . b. Find the inverse function. c. Use part b. to determine how long it takes for the car to reach a speed of 150 \(\mathrm{ft} / \mathrm{sec}\) .

Step-by-Step Solution

Answer

a. 116.5 ft/s; b. Inverse: \( t = \frac{4(v - 54)}{25} \); c. 15.36 sec.

Step 1: Evaluate the Velocity Function at 10 seconds

To find the velocity of the car at 10 seconds, substitute \( t = 10 \) into the velocity function \( v(t) = \frac{25}{4} t + 54 \). We perform the substitution and simplify:\[v(10) = \frac{25}{4} \times 10 + 54 = \frac{250}{4} + 54 = 62.5 + 54 = 116.5 \text{ feet per second.}\]

Step 2: Find the Inverse Function

To find the inverse of the function \( v(t) = \frac{25}{4}t + 54 \), solve for \( t \) in terms of \( v \). Begin by setting \( v = \frac{25}{4}t + 54 \), then:Subtract 54 from both sides:\[ v - 54 = \frac{25}{4}t \]Multiply both sides by \( \frac{4}{25} \):\[ t = \frac{4(v - 54)}{25} \]Thus, the inverse function \( t = v^{-1}(v) \) is:\[ t = \frac{4(v - 54)}{25} \]

Step 3: Solve for the Time to Reach 150 ft/s

Using the inverse function found in Step 2, determine the time \( t \) when \( v = 150 \) feet per second.Substitute \( v = 150 \) into the inverse function:\[ t = \frac{4(150 - 54)}{25} = \frac{4 \times 96}{25} = \frac{384}{25} = 15.36 \text{ seconds} \]

Key Concepts

Understanding Inverse Functions

In calculus, inverse functions are essential because they let us switch the roles of the input and output of a given function. For example, if you have a function that maps a value of time to velocity, the inverse function will tell you the time when you know the velocity.
If a function is represented as \( y = f(x) \), its inverse is written as \( x = f^{-1}(y) \). The process involves solving for the original input variable in terms of the output. This switch helps in various applications, such as finding when certain events happen instead of calculating their outcomes.
To find an inverse function, follow these steps:

• Set the function equal to its output variable.
• Solve this equation for the original input variable.
• Switch the variables, preparing the function to offer the inverse relationship.

In the velocity function example provided, we solved \( v = \frac{25}{4}t + 54 \) for \( t \), giving us the inverse function as \( t = \frac{4(v - 54)}{25} \). Now, you can determine time \( t \) when you know a specific velocity \( v \).
This step is crucial in real-world problems, like calculating how long it takes for a car to reach a certain speed, or in any scenario where finding the time from a known output is needed.

The Role of Velocity Functions in Calculus

The concept of a velocity function is especially useful in understanding motion over time. It provides a mathematical model that describes how the speed of an object changes.
In the given example, the velocity function is defined as \( v(t) = \frac{25}{4}t + 54 \). Here, \( t \) represents time in seconds, and \( v(t) \) gives us the velocity of the car in feet per second at any time \( t \).
When a velocity function is linear, as seen here, it implies constant acceleration. The slope of the line, \( \frac{25}{4} \), represents the rate of acceleration, meaning every second, the velocity increases by \( 6.25 \) feet per second.
To use a velocity function effectively:

• Plug in specific time values to find the velocity at those times.
• Analyze the function's slope to understand acceleration.
• Use it together with inverse functions to find times from given velocities.

This approach allows us to perform calculations about motion that are central to the study of calculus and physics.

Basic Principles of Kinematics

Kinematics is a branch of mechanics focused on motion and is pivotal in fields like physics and engineering. It helps explain how objects move by focusing solely on the object's movement without considering the forces causing it.
In a calculus setting, kinematics often deals with derivatives and integrals. These help relate an object's position, velocity, and acceleration:

• Position describes where an object is at a given time.
• Velocity is the rate of change of position over time.
• Acceleration is the rate of change of velocity over time.

From a kinematics perspective, the exercise provided uses the velocity function to explore how fast a car is moving at different times and how long it will take for certain speed milestones. The function \( v(t) = \frac{25}{4}t + 54 \) gives us a clear view of the car’s velocity over time, assuming a consistent acceleration.
When solving kinematics problems:

• Start with what you know — the initial velocity, positions, or time.
• Use calculus principles, like differentiation, to find unknowns like acceleration.
• Inverse functions allow you to switch between time and velocity easily.

This strong foundation in kinematics will help tackle more complex motions and forces in physics and engineering, providing a vital understanding of how objects behave over time.