Problem 26 - College Physics

Question

The driver of a car traveling on the highway suddenly slams on the brakes because of a slowdown in traffic ahead. If the car's speed decreases at a constant rate from \(60 \mathrm{mi} / \mathrm{h}\) to \(40 \mathrm{mi} / \mathrm{h}\) in \(3 \mathrm{~s},\) (a) what is the magnitude of its acceleration, assuming that it continues to move in a straight line? (b) What distance does the car travel during the braking period? Express your answers in feet.

Step-by-Step Solution

Answer

The car's acceleration is 9.78 ft/s², and it travels 220.05 feet during braking.

Step 1: Understanding the Problem

We need to find the car's acceleration and the distance traveled. The initial speed is 60 mi/h, the final speed is 40 mi/h, and the time taken to change speed is 3 seconds. We must convert these speeds from miles per hour to feet per second to use them in our calculations.

Step 2: Convert Speeds to Feet per Second

Convert the initial speed from 60 mi/h to feet per second using the conversion factor 1 mi/h = 1.467 feet/second. Similarly, convert the final speed from 40 mi/h to feet per second. Thus:\[v_i = 60 imes 1.467 = 88.02 ext{ ft/s}\]\[v_f = 40 imes 1.467 = 58.68 ext{ ft/s}\]

Step 3: Calculate Acceleration

Using the formula for acceleration, which is the change in velocity divided by time: \(a = \frac{v_f - v_i}{t}\). Substitute the values:\[a = \frac{58.68 - 88.02}{3} = -9.78 ext{ ft/s}^2\]The acceleration is -9.78 ft/s², denoting a deceleration.

Step 4: Calculate Distance Traveled Using Average Velocity

The distance traveled can be found using the formula for distance: \(d = v_{avg} \times t\). The average velocity \(v_{avg}\) is \(\frac{v_i + v_f}{2}\):\[v_{avg} = \frac{88.02 + 58.68}{2} = 73.35 ext{ ft/s}\]Now multiply the average velocity by time:\[d = 73.35 \times 3 = 220.05 ext{ ft}\]

Step 5: Conclusion

The magnitude of the car's acceleration is 9.78 ft/s², and the car travels a distance of 220.05 feet during the braking period.

Key Concepts

Acceleration

Acceleration is a key concept in kinematics, as it represents how quickly the velocity of an object changes over time. In this exercise, when the driver slams on the brakes, the car's speed decreases from 60 mi/h to 40 mi/h. Since this occurs over a three-second period, we can calculate the magnitude of acceleration using the formula:\[a = \frac{v_f - v_i}{t}\] - Here, \(v_i\) is the initial velocity (converted to ft/s), \(v_f\) is the final velocity (also converted to ft/s), and \(t\) is the time in seconds.- Inserting the values from the problem, the acceleration is calculated as \(-9.78 \text{ ft/s}^2\). The negative sign indicates deceleration, which means the car is slowing down. Acceleration is an essential part of understanding how objects move, especially in analyzing changes in speed.

Velocity Conversion

When solving kinematics problems, especially those involving velocity, it is often necessary to convert units to ensure consistency. In this scenario, the car's speeds are initially given in miles per hour but need to be converted to feet per second for more accurate calculations:- The conversion factor used is \(1 \text{ mi/h} = 1.467 \text{ ft/s}\).- For an initial speed of 60 mi/h, the conversion is: \[ 60 \times 1.467 = 88.02 \text{ ft/s} \]- For a final speed of 40 mi/h, the conversion is: \[ 40 \times 1.467 = 58.68 \text{ ft/s} \] Using the correct units is crucial for applying formulas accurately and avoiding errors in calculations.

Distance Calculation

Calculating the distance traveled during the car's deceleration involves knowing the average velocity. The distance formula used is:\[ d = v_{avg} \times t\] - The average velocity \(v_{avg}\) is the mean of the initial and final velocities: \[ v_{avg} = \frac{v_i + v_f}{2} = \frac{88.02 + 58.68}{2} = 73.35 \text{ ft/s} \]- Subsequently, the distance \(d\) can be calculated by multiplying average velocity by the time period: \[ d = 73.35 \times 3 = 220.05 \text{ feet} \]This demonstrates how knowing the average speed and time can help determine the distance covered even during changes in speed.

Deceleration

Deceleration occurs when an object reduces its speed, resulting in negative acceleration. In this car scenario, the driver applies brakes, leading to deceleration. - The earlier calculation showed an acceleration value of \(-9.78 \text{ ft/s}^2\).- The negative value emphasizes a decrease in speed rather than an increase.Understanding deceleration is vital for analyzing stopping distances and safety in vehicles. It's also essential for real-world applications like planning safe driving strategies and designing effective braking systems.