Problem 36 - College Physics

Question

The "reaction time" of the average automobile driver is about \(0.7 \mathrm{~s}\). (The reaction time is the interval between the perception of a signal to stop and the application of the brakes.) If an automobile can slow down with an acceleration of \(12.0 \mathrm{ft} / \mathrm{s}^{2}\), compute the total distance covered in coming to a stop after a signal is observed (a) from an initial velocity of \(15.0 \mathrm{mi} / \mathrm{h}\) (in a school zone) and (b) from an initial velocity of \(55.0 \mathrm{mi} / \mathrm{h}\).

Step-by-Step Solution

Answer

(a) 35.6 ft, (b) 328.4 ft.

Step 1: Convert Initial Velocity from mph to fps

Start by converting the initial velocities given in miles per hour (mph) to feet per second (fps) since the acceleration is in feet per second squared (ft/s²). Use the conversion factor: \[ 1 ext{ mph} = \frac{5280 ext{ ft}}{3600 ext{ s}} \approx 1.467 ext{ fps} \].For (a) 15 mph: \[ 15 ext{ mph} \times 1.467 \approx 22.005 ext{ fps} \].For (b) 55 mph: \[ 55 ext{ mph} \times 1.467 \approx 80.685 ext{ fps} \].

Step 2: Calculate Distance Covered During Reaction Time

Calculate the distance each car travels during the driver's reaction time of 0.7 seconds at a constant speed.Use the formula for distance: \[ d = v \times t \].For (a) initial velocity of 22.005 fps:\[ d = 22.005 \times 0.7 \approx 15.404 \text{ ft} \].For (b) initial velocity of 80.685 fps:\[ d = 80.685 \times 0.7 \approx 56.48 \text{ ft} \].

Step 3: Calculate Braking Distance

Determine the distance required to come to a stop after the brakes are applied using the formula:\[ v^2 = 2as \], where \( v \) is the initial velocity, \( a \) is the deceleration (negative acceleration), and \( s \) is the braking distance.Rearrange for \( s \): \[ s = \frac{v^2}{2a} \].With \( a = 12 \text{ ft/s}^2 \):For (a) initial velocity of 22.005 fps:\[ s = \frac{(22.005)^2}{2 \times 12} \approx 20.18 \text{ ft} \].For (b) initial velocity of 80.685 fps:\[ s = \frac{(80.685)^2}{2 \times 12} \approx 271.89 \text{ ft} \].

Step 4: Calculate Total Stopping Distance

Add the distance covered during the reaction time to the braking distance for the total stopping distance.For (a):\[ 15.404 \text{ ft during reaction} + 20.18 \text{ ft braking} = 35.584 \text{ ft total} \].For (b):\[ 56.48 \text{ ft during reaction} + 271.89 \text{ ft braking} = 328.37 \text{ ft total} \].

Key Concepts

Reaction Time Calculation

Reaction time is a crucial factor in understanding how a driver responds to a signal to stop. It refers to the interval between noticing the signal and actually applying the brakes. For the average driver, this time is around 0.7 seconds. During this short span, the car continues to travel at its initial velocity, which means the vehicle covers some distance before the brakes even engage.

In this exercise, we calculate the distance traveled during this reaction time by using the formula: \[ d = v \times t \]where \( d \) is the distance, \( v \) is the velocity, and \( t \) is the reaction time. This formula shows us how far a car will travel if it continues moving at a constant speed over a short period. By remembering this simple concept and formula, you can accurately determine how much ground is covered before any braking begins.

Velocity Conversion

Converting velocity from miles per hour (mph) to feet per second (fps) is a common task in physics, especially when working with different units of measurement. This conversion is essential because acceleration is often given in feet per second squared (ft/s²), so our initial velocity should match those units.

To convert mph to fps, use the conversion factor:\[ 1 \text{ mph} = \frac{5280 \text{ ft}}{3600 \text{ s}} \approx 1.467 \text{ fps} \]By multiplying the velocity in mph by this conversion factor, you can easily transform it into fps. For example:

• 15 mph converts to \( 15 \times 1.467 = 22.005 \text{ fps} \)
• 55 mph converts to \( 55 \times 1.467 = 80.685 \text{ fps} \)

This conversion allows us to work consistently with other given measurements, such as acceleration or braking distances.

Braking Distance

Braking distance is the distance a vehicle covers from the time the brakes are applied until it comes to a complete stop. This distance depends on several factors, including the speed of the vehicle and the rate of deceleration, which is provided in this exercise as feet per second squared.

We calculate braking distance using the equation:\[ v^2 = 2a s \]where \( v \) is the initial velocity, \( a \) is the deceleration (negative of acceleration), and \( s \) is the braking distance. Solving for \( s \), we get:\[ s = \frac{v^2}{2a} \]

• For an initial velocity of 22.005 fps with an acceleration of 12 ft/s², the distance is \( \frac{(22.005)^2}{2 \times 12} \approx 20.18 \text{ ft} \).
• At 80.685 fps, the distance extends to \( \frac{(80.685)^2}{2 \times 12} \approx 271.89 \text{ ft} \).

Understanding braking distance helps us appreciate how different speeds affect the distance needed to stop, emphasizing the importance of safe driving speeds.

Stopping Distance

Stopping distance is the total distance a vehicle travels before it comes to a complete stop after observing a signal. It's the sum of two main components: the distance covered during the driver's reaction time and the braking distance.

To find the total stopping distance, simply add the calculated reaction distance and the braking distance:

• For an initial velocity of 22.005 fps: Reaction distance is 15.404 ft, and braking distance is 20.18 ft, giving a total of \( 15.404 + 20.18 = 35.584 \text{ ft} \).
• With an initial velocity of 80.685 fps: The distances are 56.48 ft during the reaction time and 271.89 ft while braking, totaling \( 56.48 + 271.89 = 328.37 \text{ ft} \).

This comprehensive understanding of stopping distance underscores the significance of alertness and adequate spacing between vehicles, facilitating better decision-making while driving.