Problem 37 - College Physics

Question

According to recent typical test data, a Ford Focus travels \(0.250 \mathrm{mi}\) in \(19.9 \mathrm{~s},\) starting from rest. The same car, when braking from \(60.0 \mathrm{mi} / \mathrm{h}\) on dry pavement, stops in \(146 \mathrm{ft}\). Assume constant acceleration in each part of its motion, but not necessarily the same acceleration when slowing down as when speeding up. (a) Find this car's acceleration while braking and while speeding up. (b) If its acceleration is constant while speeding up, how fast (in \(\mathrm{mi} / \mathrm{h}\) ) will the car be traveling after \(0.250 \mathrm{mi}\) of acceleration? (c) How long does it take the car to stop while braking from \(60.0 \mathrm{mi} / \mathrm{h} ?\)

Step-by-Step Solution

Answer

(a) Speeding up: \( a = 7.98 \; \mathrm{ft/s}^2 \), braking: \( a = -28.2 \; \mathrm{ft/s}^2 \). (b) 54.4 mi/h. (c) 8.87 seconds.

Step 1: Understanding the Problem

The problem involves finding acceleration during both speeding up and braking phases. We use kinematic equations assuming constant acceleration. Then, we'll convert the found quantities into necessary units for comparison.

Step 2: Convert Units

We need all measurements in consistent units. Convert 0.250 miles to feet by multiplying by 5280 feet/mile. Convert 60.0 mi/h to feet/second by multiplying by \( \frac{5280}{3600} \).

Step 3: Acceleration During Speeding Up

Use the kinematic equation \( s = ut + \frac{1}{2}at^2 \) where \( u = 0 \), \( s = 0.250 \times 5280 \), and \( t = 19.9 \). Solve for \( a \).

Step 4: Acceleration While Braking

Use the equation \( v^2 = u^2 + 2as \), where \( v = 0 \), \( u = 60 \times \frac{5280}{3600} \), and \( s = 146 \). Solve for \( a \).

Step 5: Speed After Acceleration

Use the equation \( v = u + at \) to find the final speed \( v \), using \( u = 0 \), \( a \) from speeding up, and \( t = 19.9 \). Convert \( v \) to mi/h.

Step 6: Time to Stop While Braking

Use the equation \( v = u + at \) with \( v = 0 \), initial speed \( u \) from step 4, and solve for \( t \) using \( a \) from braking.

Key Concepts

Acceleration Calculation

Acceleration is the rate at which an object's velocity changes over time. In this exercise, we need to calculate acceleration for a car during two different parts of its journey: while speeding up and while braking. To find acceleration, we use kinematic equations. These equations relate speed, distance, time, and acceleration.

• For speeding up, we can use the equation: \( s = ut + \frac{1}{2}at^2 \).
• For braking, the equation: \( v^2 = u^2 + 2as \) can be applied.

Remember, the first equation helps when you know the time, and the second is handy when time is not part of the data given. Always identify whether the initial velocity is zero or not. Here, the car starts from rest when speeding up and stops while braking. Grasping these equations allows you to determine how acceleration impacts motion. Understanding this is fundamental for both parts of the problem.

Kinematic Equations

Kinematic equations are vital for analyzing motion in physics, especially with constant acceleration. The main equations include:

• \( v = u + at \)
• \( s = ut + \frac{1}{2}at^2 \)
• \( v^2 = u^2 + 2as \)
• \( s = vt - \frac{1}{2}at^2 \)

These equations allow us to link displacement \( s \), initial velocity \( u \), final velocity \( v \), acceleration \( a \), and time \( t \). In the exercise, we use these equations to track how the Ford Focus accelerates and brakes. By rearranging these equations, you can solve for different variables depending on what you need. For example, if you know the distance and initial velocity, you can calculate acceleration or time. The exercise showcases their usefulness in real-world scenarios. Whether you are learning about acceleration or predicting motion, kinematic equations are the key tools in your physics toolbox.

Unit Conversion

Unit conversion is essential, especially when analyzing problems in physics where units must be consistent. Often, we encounter different units, like miles, feet, or meters. You must convert these units into a standard form. For example:

• To convert miles to feet, use the fact that 1 mile equals 5280 feet.
• To convert speeds from miles per hour to feet per second, multiply the speed by \( \frac{5280}{3600} \).

These conversions make equations work correctly, as the kinematic equations need coherent units. In the exercise, converting distances and speeds was crucial. Without these conversions, the calculations for acceleration and time would not make sense.Unit conversion bridges the gap between different measurement systems, allowing accurate calculations in physics. It's a straightforward but crucial step that ensures your analysis remains valid. Always check your units before diving into any problem to avoid mistakes.

Motion Analysis

Motion analysis involves examining how objects move and how forces like acceleration affect them. For our exercise, motion analysis ties together our calculations and understandings from kinematics, stress on constant acceleration, and unit conversion. When analyzing the car's motion:

• Determine initial conditions: starting from rest when speeding, stopping when braking.
• Use kinematic equations to calculate how quickly the car accelerates and decelerates.
• Check the converted units to make sure they are correctly applied in your calculations.

Breaking the motion into segments (speeding up and braking) allows focusing on different forces in each part. It's not enough to just calculate numbers; understanding what those numbers represent in terms of motion is key. Motion Analysis helps visualize the path taken, the speed changes, and how those changes occur over time. This approach is essential for practical physics applications, like designing safer cars or understanding traffic flow.