Problem 50 - Essential Calculus Early Transcendentals

Question

A car is traveling at 50 \(\mathrm{mi} / \mathrm{h}\) when the brakes are fully applied, producing a constant deceleration of 22 \(\mathrm{ft} / \mathrm{s}^{2} .\) What is the distance traveled before the car comes to a stop?

Step-by-Step Solution

Answer

The car travels approximately 122.2 feet before stopping.

Step 1: Convert Initial Speed to Feet per Second

The car's initial speed is given as 50 miles per hour. We need to convert this to feet per second to match the unit of deceleration. We know:1 mile = 5280 feet and 1 hour = 3600 seconds.Therefore, convert speed: \[ 50 \text{ miles/hour} = 50 \times \frac{5280}{3600} \text{ ft/s} \approx 73.33 \text{ ft/s} \]

Step 2: Use Kinematic Equation for Final Speed

The final speed (v) of the car when it stops is 0 ft/s. We need to find the distance using the kinematic equation:\[ v^2 = u^2 + 2as \]where: \( u = 73.33 \text{ ft/s} \) is the initial speed, \( v = 0 \text{ ft/s} \) is the final speed, \( a = -22 \text{ ft/s}^2 \) is the deceleration (negative because it is slowing down), and \( s \) is the distance we want to find.

Step 3: Solve for Distance

Rearrange the equation \( v^2 = u^2 + 2as \) to solve for \( s \):\[ s = \frac{v^2 - u^2}{2a} \]Substitute the known values:\[ s = \frac{0^2 - (73.33)^2}{2(-22)} \]Calculate \((73.33)^2\):\[ 73.33^2 = 5376.8889 \]Then,\[ s = \frac{-5376.8889}{-44} \approx 122.2 \text{ feet} \]

Key Concepts

Deceleration

Deceleration occurs when an object slows down, which means its velocity decreases over time. In our exercise, the car experiences a constant deceleration of 22 \( \text{ft/s}^2 \). This value is negative because it opposes the direction of the car's initial velocity.
To understand deceleration fully, consider it as the opposite of acceleration. While acceleration increases the speed of an object, deceleration reduces it.
This concept is crucial when solving problems related to stopping distances, as deceleration directly influences how quickly a moving object can be brought to rest.

• Deceleration is measured in terms of units of velocity per time squared (e.g., \( \text{ft/s}^2 \)).
• A larger deceleration value signifies a faster stop.
• Deceleration always has a negative effect when calculating changes in velocity.

Understanding deceleration can help us predict how long a moving object will take to come to a complete stop.

Unit Conversion

Unit conversion is the process of converting one unit of measurement into another, which is crucial for consistent calculations. In this exercise, we converted speed from miles per hour (mi/h) to feet per second (ft/s) because the deceleration was provided in ft/s².
Here's how you can convert speed from mi/h to ft/s:

• Know the conversion factors:
  \( 1 \text{ mile} = 5280 \text{ feet} \)
  \( 1 \text{ hour} = 3600 \text{ seconds} \)
• To convert 50 mi/h to ft/s, use the formula:
  \[ 50 \left(\frac{5280}{3600}\right) \text{ ft/s} \approx 73.33 \text{ ft/s} \]

Always ensure your units are consistent. Using mismatched units can lead to incorrect results when applying formulas, especially in physics problems where precise calculation is key.

Distance Calculation

Distance calculation using kinematic equations involves applying established formulas to solve motion-related problems. In our example, we used the kinematic equation \( v^2 = u^2 + 2as \) to determine the distance a car travels before stopping.
Here, each variable represents:

• \( v \): Final velocity (0 ft/s since the car stops)
• \( u \): Initial velocity (73.33 ft/s, post-conversion)
• \( a \): Acceleration (or deceleration, -22 ft/s² in this case)
• \( s \): Distance to calculate

We rearrange the equation to solve for \( s \): \( s = \frac{v^2 - u^2}{2a} \). Substituting the known values, we get:
\[ s = \frac{0^2 - (73.33)^2}{2(-22)} = 122.2 \text{ feet} \]
This equation allows us to compute how far the car travels during deceleration, emphasizing the importance of having all parameters in consistent units for accurate computation.