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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.

Short Answer

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

Step by step solution

01

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.
02

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:\[v_i = 60 imes 1.467 = 88.02 ext{ ft/s}\]\[v_f = 40 imes 1.467 = 58.68 ext{ ft/s}\]
03

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.
04

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:\[d = 73.35 \times 3 = 220.05 ext{ ft}\]
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

While riding on a bus traveling down the highway, you notice that it takes 2 min to travel from one roadside mile marker to the next one. (a) What is your speed in \(\mathrm{mi} / \mathrm{h} ?\) (b) How long does it take the bus to travel 100 yds?

A hot-air balloonist, rising vertically with a constant speed of \(5.00 \mathrm{~m} / \mathrm{s},\) releases a sandbag at the instant the balloon is \(40.0 \mathrm{~m}\) above the ground. (See Figure \(2.54 .)\) After it is released, the sandbag encounters no appreciable air drag. (a) Compute the position and velocity of the sandbag at \(0.250 \mathrm{~s}\) and \(1.00 \mathrm{~s}\) after its release. (b) How many seconds after its release will the bag strike the ground? (c) How fast is it moving as it strikes the ground? (d) What is the greatest height above the ground that the sandbag reaches? (e) Sketch graphs of this bag's acceleration, velocity, and vertical position as functions of time.

Two bicyclists start a sprint from rest, each riding with a constant acceleration. Bicyclist \(A\) has twice the acceleration of bicyclist \(B ;\) however, bicyclist \(B\) rides for twice as long as bicyclist \(A\). What is the ratio of the distance traveled by bicyclist \(A\) to that traveled by bicyclist \(B\) ? What is the ratio of the speed of bicyclist \(A\) to that of bicyclist \(B\) at the end of their sprint?

A helicopter 8.50 m above the ground and descending at \(3.50 \mathrm{~m} / \mathrm{s}\) drops a package from rest (relative to the helicopter). Just as it hits the ground, find (a) the velocity of the package relative to the helicopter and (b) the velocity of the helicopter relative to the package. The package falls freely.

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 \(60.0 \mathrm{mi} / \mathrm{h} ?\)

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