Problem 13 - Basic Technical Mathematics with Calculus

Question

A certain Chevrolet Corvette goes from 0 mi/h to \(60.0 \mathrm{mi} / \mathrm{h}\) \((88.0 \mathrm{ft} / \mathrm{s})\) in \(3.60 \mathrm{s}\). Assuming constant acceleration, how far (in \(\mathrm{ft}\) ) does it travel in this time?

Step-by-Step Solution

Answer

The Corvette travels approximately 159 ft in 3.60 seconds.

Step 1: Identify the Known Variables

We are given the final velocity \( v = 88.0 \, \mathrm{ft/s} \), initial velocity \( u = 0 \, \mathrm{ft/s} \), and time \( t = 3.60 \, \mathrm{s} \). Our task is to find the distance \( s \), assuming constant acceleration.

Step 2: Use the Equation of Motion

The equation that relates initial velocity, final velocity, time, and distance with constant acceleration is \( v = u + at \). First, we will use this equation to find acceleration \( a \).

Step 3: Calculate Acceleration

Rearrange the equation to solve for acceleration: \( a = \frac{v - u}{t} \). Substitute the known values: \( a = \frac{88.0 - 0}{3.60} = 24.44 \, \mathrm{ft/s^2} \).

Step 4: Utilize the Distance Formula

With acceleration known, use the equation \( s = ut + \frac{1}{2} a t^2 \) to find the distance \( s \) traveled. Since initial velocity \( u = 0 \), the equation simplifies to \( s = \frac{1}{2} a t^2 \).

Step 5: Calculate the Distance

Substitute the values into the equation: \( s = \frac{1}{2} \times 24.44 \times (3.60)^2 \). Calculate to find \( s = 158.98 \, \mathrm{ft} \).

Step 6: Round the Result

Convert the calculated distance into a rounded value. So, the distance traveled is approximately \( 159 \, \mathrm{ft} \).

Key Concepts

Constant Acceleration

Constant acceleration means that the rate of change of velocity is steady over time. In other words, the acceleration does not vary as an object moves. This concept is crucial in the study of physics since it allows us to apply several consistent formulas to predict the motion of objects. Consider the example of a car speeding up from a stoplight. If the car experiences constant acceleration, this means that in each passing second, it gains the same amount of speed. This simplification helps us use mathematical equations to find out other characteristics of the motion, like how far the car travels in a given time. In our exercise, the Chevrolet Corvette goes from 0 to 60 mph under constant acceleration. Using this assumption of constant rate, we can easily calculate other properties, such as distance traveled or the acceleration rate itself.

Equations of Motion

The equations of motion are set rules that predict the behavior of moving objects under constant acceleration. These equations help us find unknown values when certain variables like initial velocity, final velocity, time, and distance are known. There are three main equations of motion:

• The first equation relates velocity, acceleration, and time: \( v = u + at \)
• The second equation connects distance, initial velocity, time, and acceleration: \( s = ut + \frac{1}{2} at^2 \)
• The third equation provides a relationship between velocities and distance: \( v^2 = u^2 + 2as \)

In the given problem, we use the first two equations. First, we determine acceleration using the initial and final velocities and the time. Then, we use the known acceleration in the second equation to find the distance traveled.

Kinematics

Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause them to move. It focuses purely on measuring and describing motion. The key elements involved in kinematic studies are:

• Position: Where an object is located in space
• Velocity: The speed and direction in which an object moves
• Acceleration: The rate of change of velocity over time

In our exercise, the Corvette's movement has been described by its initial and final velocities, exhibiting the principles of kinematics. We use its known velocities and time to calculate acceleration, and then proceed to determine the distance it travels using these kinematic equations.

Distance Calculation

Calculating distance when an object is under constant acceleration involves knowing the initial velocity, acceleration, and time. The formula used in our exercise is:\[ s = ut + \frac{1}{2} at^2 \]This formula stems from the concept that if you accelerate from an initial speed, the distance traveled over time consists of what is covered initially plus the additional distance from accelerating. In the current exercise, the initial velocity is zero, simplifying the formula to:\[ s = \frac{1}{2} at^2 \]By substituting the values obtained from our acceleration calculation and time (\( t = 3.60 \, s \)), we compute the total distance traveled by the car during its acceleration phase. This method of distance calculation is fundamental in physics problems involving kinematics and it provides a clear view into the effectiveness of consistent acceleration over time.