Problem 36 - College Physics

Question

A curving freeway exit has a radius of 50.0 \(\mathrm{m}\) and a posted speed limit of 35 \(\mathrm{mi} / \mathrm{h} .\) What is your radial acceleration (in \(\mathrm{m} /\mathrm{s}^{2} )\) if you take this exit at the posted speed? What if you take the exit at a speed of 50 \(\mathrm{mi} / \mathrm{h} ?\)

Step-by-Step Solution

Answer

At 35 mi/h, the radial acceleration is 4.90 m/s², and at 50 mi/h, it is 9.99 m/s².

Step 1: Convert Speed from Miles per Hour to Meters per Second

To calculate the radial acceleration, we need to have the speed in meters per second. The conversion factor is: 1 mile per hour = 0.44704 meters per second.First, convert 35 mi/h:\[35 \text{ mi/h} \times 0.44704 \frac{\text{m/s}}{\text{mi/h}} = 15.6464 \text{ m/s}\]Next, convert 50 mi/h:\[50 \text{ mi/h} \times 0.44704 \frac{\text{m/s}}{\text{mi/h}} = 22.352 \text{ m/s}\]

Step 2: Calculate Radial Acceleration at 35 mi/h

Radial acceleration can be calculated using the formula:\[a_r = \frac{v^2}{r}\]Where:- \(a_r\) is the radial acceleration,- \(v\) is the speed (15.6464 m/s in this case), and- \(r\) is the radius of the curve (50.0 m).Substitute the values:\[a_r = \frac{(15.6464)^2}{50.0} = \frac{244.8}{50.0} = 4.896 \text{ m/s}^2\]

Step 3: Calculate Radial Acceleration at 50 mi/h

Using the same formula for radial acceleration:\[a_r = \frac{v^2}{r}\]Substitute the speed 22.352 m/s:\[a_r = \frac{(22.352)^2}{50.0} = \frac{499.422}{50.0} = 9.9884 \text{ m/s}^2\]

Key Concepts

Unit Conversion

When dealing with problems that involve speed and radial acceleration, unit conversion becomes crucial. In physics, different systems of measurement can lead to confusion if not properly managed. Here, speed was initially given in miles per hour (mi/h). However, to calculate radial acceleration, we need speed in meters per second (m/s). To perform this conversion, we use the relation:

• 1 mile per hour is equivalent to 0.44704 meters per second.

By using this conversion factor, we change 35 mi/h into meters per second: 35 mi/h * 0.44704 m/s/mi/h = 15.6464 m/s. Similarly, for 50 mi/h, the conversion is: 50 mi/h * 0.44704 m/s/mi/h = 22.352 m/s. Converting units accurately ensures that calculations performed using standard formulas are correct. Without correct units, even a small error in conversion can lead to significantly wrong results.

Circular Motion

Circular motion involves any object moving in a circular path. In such motion, objects experience a type of acceleration known as radial or centripetal acceleration. This acceleration is always directed towards the center of the circular path.For circular motion:

• The path's radius is a key component, affecting the magnitude of the radial acceleration.
• The speed of the object also influences radial acceleration considerably.

Let's consider a car on a curved freeway exit, maintaining a constant speed. The car will always have radial acceleration because its direction is continually changing, even if its speed remains constant.The formula used, \(a_r = \frac{v^2}{r}\), represents radial acceleration. Here:

• \(a_r\) is radial acceleration,
• \(v\) is the speed of the object,
• and \(r\) is the radius of the circle.

Understanding these relationships helps predict how changes in speed or radius will affect the object's behavior on its path.

Speed Calculation

Calculating speed and its impact on radial acceleration is fundamental when analyzing circular motion. Speed, in this context, isn't just about how fast an object moves but is pivotal in determining acceleration.Consider the posted speed limit and your actual speed on a curved exit. Following the posted speed limit of 35 mi/h:

• After converting, we find 15.6464 m/s.
• Using the formula \(a_r = \frac{v^2}{r}\), the calculated radial acceleration is approximately 4.896 m/s².

However, if you increase your speed to 50 mi/h:

• Converted into meters per second, this is 22.352 m/s.
• Applying the same formula, the radial acceleration jumps to about 9.9884 m/s².

Speed, therefore, greatly influences how much radial acceleration an object experiences. Doubling speed doesn't just double the acceleration due to the squared relationship in the formula, highlighting the exponential effect of speed on acceleration in circular paths.