Problem 30 - College Physics

Question

A car is traveling at \(60 \mathrm{mi} / \mathrm{h}\) down a highway. (a) What magnitude of acceleration does it need to have to come to a complete stop in a distance of \(200 \mathrm{ft} ?\) (b) What acceleration does it need to stop in \(200 \mathrm{ft}\) if it is traveling at \(100 \mathrm{mi} / \mathrm{h} ?\)

Step-by-Step Solution

Answer

(a) Acceleration = -19.36 ft/s²; (b) Acceleration = -53.80 ft/s².

Step 1: Understand the Problem

We have a car traveling at a certain speed and we want it to stop completely over a given distance. For this, we will use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance. This equation is \( v_f^2 = v_i^2 + 2a d \), where \( v_f \) is the final velocity (0, since the car stops), \( v_i \) is the initial velocity, \( a \) is the acceleration, and \( d \) is the distance over which the car stops.

Step 2: Convert Units (Part a)

First, convert the initial speed from miles per hour to feet per second because acceleration will be calculated in \( \text{ft/s}^2 \).\[60 \ \text{mi/h} = \frac{60 \times 5280}{3600} \ \text{ft/s} = 88 \ \text{ft/s} \].

Step 3: Apply the Kinematic Equation (Part a)

Now, use the equation with \( v_f = 0 \), \( v_i = 88 \ \text{ft/s} \), and \( d = 200 \ \text{ft} \).\[ 0 = 88^2 + 2a \times 200 \]Solve for \( a \):\[ 0 = 7744 + 400a \]\[ -7744 = 400a \]\[ a = \frac{-7744}{400} = -19.36 \ \text{ft/s}^2 \].

Step 4: Convert Units (Part b)

Convert the initial speed from miles per hour to feet per second for the second part.\[100 \ \text{mi/h} = \frac{100 \times 5280}{3600} \ \text{ft/s} = 146.67 \ \text{ft/s} \].

Step 5: Apply the Kinematic Equation (Part b)

Use \( v_f = 0 \), \( v_i = 146.67 \ \text{ft/s} \), and \( d = 200 \ \text{ft} \) in the equation:\[ 0 = (146.67)^2 + 2a \times 200 \]Solve for \( a \):\[ 0 = 21520.89 + 400a \]\[ -21520.89 = 400a \]\[ a = \frac{-21520.89}{400} = -53.80 \ \text{ft/s}^2 \].

Key Concepts

Equations of Motion

One of the backbone ideas in kinematics is the equations of motion. These equations help predict the future behavior of moving objects. This is due to their relationship between factors such as velocity, acceleration, and displacement. In our case, the equation in focus is \( v_f^2 = v_i^2 + 2ad \). This relates to:

• \( v_f \): Final velocity. This is the speed of the object when it reaches its destination or stops. For our car problem, the car comes to a stop, so \( v_f = 0 \).
• \( v_i \): Initial velocity. The starting speed of the car. In the exercise, these were given as \( 60 \) mph and \( 100 \) mph respectively.
• \( a \): Acceleration. It's what we are trying to find. Acceleration tells how fast or slow an object changes its speed.
• \( d \): Distance. The total area over which the car decelerates, given as \( 200 \) ft in the exercise.

Using this equation allows us to set up relationships and solve for unknown quantities like acceleration, by plugging in other known values.

Acceleration

Acceleration refers to how the velocity of an object changes with time. It is a key concept in solving kinematic problems. In the given problem, we calculate acceleration needed for the car to come to a complete stop. Since the car is stopping, our acceleration value should be negative, indicating a reduction in speed, or deceleration.

• Units: Often expressed in \( \text{ft/s}^2 \) or \( \text{m/s}^2 \), indicating how velocity changes over each second.
• Direction of Acceleration: If acceleration is negative, the object is slowing down. Conversely, if it’s positive, the object speeds up.

In the calculations:- For \( 60 \) mph: \( a = -19.36 \ \text{ft/s}^2 \)- For \( 100 \) mph: \( a = -53.80 \ \text{ft/s}^2 \)
Both results are negative, as expected since the car stops.

Unit Conversion

Unit conversion is crucial in solving physics problems, ensuring that all quantities are in compatible units before performing calculations. The initial speeds given in miles per hour (mph) need to be converted to feet per second (ft/s) since the distance is provided in feet. The conversion involves:

• 1 mile = 5280 feet
• 1 hour = 3600 seconds

The conversion formula is \( \text{ft/s} = \left( \text{mph} \times \frac{5280}{3600} \right) \).
For instance:- \( 60 \) mph = \( 88 \ \text{ft/s} \)- \( 100 \) mph = \( 146.67 \ \text{ft/s} \)
Using the correct units is vital for applying the equations effectively and ensuring the accuracy of the calculations.

Physics Problems

Tackling physics problems, like the one given, often involves a sequence of logical steps. It requires a solid understanding of the basic concepts and careful problem-solving techniques:

• Understanding the Problem: Start by identifying what you need to find. In this exercise, determining the required acceleration for the car to stop.
• Unit Consistency: Ensure all measurements are in standard units. Here, speeds are converted to \( \text{ft/s} \).
• Applying Equations: Choose an appropriate formula that relates the known and unknown quantities.
• Solving Mathematically: Carefully solve the equations algebraically to find the desired quantity.

Following these steps makes physics less daunting, transforming complex scenarios into manageable equations. Practice enhances the ability to quickly recognize what needs to be done to find solutions.