/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 76 Car \(A\) is traveling at 40 mil... [FREE SOLUTION] | 91Ó°ÊÓ

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Car \(A\) is traveling at 40 milh when it enters a 30 milh speed zone. The driver of car \(A\) decelerates at a rate of \(16 \mathrm{ft} / \mathrm{s}^{2}\) until reaching a speed of \(30 \mathrm{mil}\), which she then maintains. When car \(B\), which was initially \(60 \mathrm{ft}\) behind car \(A\) and traveling at a constant speed of \(45 \mathrm{mil}\), enters the speed zone, its driver decelerates at a rate of \(20 \mathrm{ft}\) s " until reaching a speed of \(28 \mathrm{mi} / \mathrm{h}\). Knowing that the driver of car \(B\) maintains a speed of \(28 \mathrm{mi} / \mathrm{h}\), determine \((a)\) the closest that car \(B\) comes to car \(A,(b)\) the time at which car \(A\) is \(70 \mathrm{ft}\) in front of car \(B .\)

Short Answer

Expert verified
Closest distance: 52.13 ft; Time for 70-ft gap: 6.09 seconds.

Step by step solution

01

Convert Speeds to Consistent Units

To solve this problem, we first need to convert the speeds from miles per hour (mph) to feet per second (ft/s).- Speed of car A before deceleration: \(40 \text{ mph} = 40 \times \frac{5280}{3600} \approx 58.67 \text{ ft/s}\)- Speed of car A after deceleration: \(30 \text{ mph} = 30 \times \frac{5280}{3600} = 44 \text{ ft/s}\)- Speed of car B before deceleration: \(45 \text{ mph} = 45 \times \frac{5280}{3600} \approx 66 \text{ ft/s}\)- Speed of car B after deceleration: \(28 \text{ mph} = 28 \times \frac{5280}{3600} \approx 41.07 \text{ ft/s}\)
02

Calculate Time and Distance for Car A to Decelerate

Car A decelerates from 58.67 ft/s to 44 ft/s at a deceleration of \(16 \text{ ft/s}^2\).- Time to decelerate (\(t_A\)) is calculated using: \(v_f = v_i - a \cdot t \Rightarrow 44 = 58.67 - 16 \cdot t_A \) Solving for \(t_A\), we get: \(t_A = \frac{58.67 - 44}{16} = 0.91 \text{ s}\)- Distance traveled during deceleration (\(d_A\)) is calculated using: \(d = v_i \cdot t + \frac{1}{2} a \cdot t^2 \) \(d_A = 58.67 \cdot 0.91 + \frac{1}{2} \cdot (-16) \cdot (0.91)^2 \approx 53.33 \text{ ft}\)
03

Calculate Time and Distance for Car B to Decelerate

Car B decelerates from 66 ft/s to 41.07 ft/s at a deceleration of \(20 \text{ ft/s}^2\).- Time to decelerate (\(t_B\)) is calculated using: \(v_f = v_i - a \cdot t \Rightarrow 41.07 = 66 - 20 \cdot t_B \) Solving for \(t_B\), we get: \(t_B = \frac{66 - 41.07}{20} = 1.20 \text{ s}\)- Distance traveled during deceleration (\(d_B\)) is calculated using: \(d = v_i \cdot t + \frac{1}{2} a \cdot t^2 \) \(d_B = 66 \cdot 1.20 - \frac{1}{2} \cdot 20 \cdot (1.20)^2 \approx 61.20 \text{ ft}\)
04

Calculate Relative Position and Distance Dynamics After Deceleration

Find the closest distance between the cars during the deceleration phase: - Initial gap between cars is 60 ft. - At the end of their deceleration, car A has traveled 53.33 ft and car B has traveled 61.20 ft. - Gap change = Distance car B travels - Initial gap - Distance car A travels. - Closest distance after deceleration = 60 + 53.33 - 61.20 = 52.13 ft. Thus, car B is 52.13 ft behind car A when both finish decelerating.
05

Determine Time When Car A is 70 ft Ahead of Car B

After deceleration, car A travels at 44 ft/s and car B at 41.07 ft/s.- The initial gap is now 52.13 ft.- Calculate the time required for a 70-ft gap: The speed difference \(v_{diff} = 44 - 41.07 = 2.93 \text{ ft/s}\).- To achieve a 70-ft gap: \(70 = 52.13 + 2.93 \times t_{gap}\).- Solving for \(t_{gap}\), we get \(t_{gap} = \frac{70 - 52.13}{2.93} \approx 6.09 \text{ s}\).
06

Final Answer Composition

-(A) The closest that car B comes to car A is 52.13 ft. -(B) The time at which car A is 70 ft in front of car B is approximately 6.09 seconds after both finish decelerating.

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Key Concepts

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

Kinematics
Kinematics is a branch of mechanics that studies the motion of objects without considering the causes of that motion. It's all about understanding how objects move, which includes analyzing variables like speed, velocity, acceleration, and displacement. In our problem with cars A and B, kinematics helps us figure out how far and how fast each car moves during and after deceleration.

To solve kinematics problems effectively, we need to convert all units to be consistent. This is why we transformed speeds from miles per hour to feet per second. By doing this, calculations become easier and more accurate.
  • Distance traveled by an object can be calculated using the formula: \[ d = v_i \cdot t + \frac{1}{2} a \cdot t^2\]where:
    • \( d \) is distance
    • \( v_i \) is initial velocity
    • \( a \) is acceleration
    • \( t \) is time
  • Changes in velocity over time can be calculated using:\[ v_f = v_i - a \cdot t\]where:
    • \( v_f \) is final velocity
Understanding these kinematic equations allows us to know how each car in the exercise moves over time, letting us predict future motion steps.
Deceleration
Deceleration is a type of acceleration where an object reduces its speed. It's the opposite of acceleration, leading to slower speeds over time. In dynamics, it's crucial to distinguish between positive acceleration (speeding up) and deceleration (slowing down).

In our scenario, both cars are slowing down to reach their respective final speeds. Car A decelerates from 40 mph to 30 mph, while car B reduces its speed from 45 mph to 28 mph. Using the formula for deceleration:\[ t = \frac{v_f - v_i}{a}\]we can find the time each car takes to decelerate. This is possible since we know both the initial and final velocities, along with the deceleration rate. The negative sign in the acceleration indicates that the object is slowing down.
  • Car A's deceleration is \(-16 \text{ ft/s}^2\).
  • Car B's deceleration is \(-20 \text{ ft/s}^2\).
By plugging in these values, together with the speeds, we find out how quickly each car reaches the desired speed. This helps us predict not only their speeds at different points in their journey but also how much distance they cover while decelerating.
Relative Motion
Relative motion examines how one object's movement is observed from the viewpoint of another. It's a fundamental part of dynamics, making us think about one object's position or speed in relation to another object's position. When dealing with two moving cars, such as in our exercise, understanding relative motion helps us see how they interact over time.

Initially, car B is 60 feet behind car A. As both cars decelerate and continue driving at constant, yet slower speeds, we can calculate their relative positions using:
  • The distance each car travels during deceleration.
  • Their speeds after deceleration.
  • The gap between them afterward.
By knowing the final positions and speeds, we can determine how long it will take for situations to change, like when car A ends up being 70 feet ahead of car B. This requires an understanding of both their speeds and the dynamic distance between them.

The key is to look at their speed difference. In this scenario, the relative speed (difference between their two speeds) shows how fast one car is moving compared to the other. This information aids in predicting when specific conditions will occur, such as the desired 70 feet gap, thus illustrating the power of relative motion in dynamics.

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Most popular questions from this chapter

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