/*! 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 28 Entering the Freeway. A car sits... [FREE SOLUTION] | 91Ó°ÊÓ

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Entering the Freeway. A car sits in an entrance ramp to a freeway, waiting for a break in the traffic. The driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 20 \(\mathrm{m} / \mathrm{s}(45 \mathrm{mi} / \mathrm{h})\) when it reaches the end of the \(120-\mathrm{m}\) -long ramp. (a) What is the acceleration of the car? (b) How much time does it take the car to travel the length of the ramp? (c) The traffic on the freeway is moving at a constant speed of 20 \(\mathrm{m} / \mathrm{s} .\) What distance does the traffic travel while the car is moving the length of the ramp?

Short Answer

Expert verified
(a) Acceleration is \(\frac{5}{3}\, \text{m/s}^2\). (b) Time is 12 seconds. (c) Traffic travels 240 m.

Step by step solution

01

Understanding the Problem

We have a car accelerating from rest with constant acceleration along a ramp, reaching a speed of 20 m/s at the end of a 120 m ramp. We need to find: (a) the car's acceleration, (b) the time taken to cover the ramp, and (c) the distance traveled by traffic on the freeway at a constant speed of 20 m/s.
02

Using Kinematic Equation for Acceleration

(a) To find acceleration, we can use the kinematic equation: \(v^2 = u^2 + 2as\), where \(v = 20\, \text{m/s}\) (final velocity), \(u = 0\, \text{m/s}\) (initial velocity), and \(s = 120\, \text{m}\) (distance). Solving for \(a\), we have: \[20^2 = 0^2 + 2a \times 120 \rightarrow a = \frac{400}{240} = \frac{5}{3}\, \text{m/s}^2\].
03

Calculating Time Taken on the Ramp

(b) To find the time taken, use the kinematic equation: \(v = u + at\). We already have \(v = 20\, \text{m/s}, u = 0\, \text{m/s},\) and \(a = \frac{5}{3}\, \text{m/s}^2\). Solving for \(t\), we find \(t = \frac{20}{\frac{5}{3}} = \frac{20 \times 3}{5} = 12\, \text{seconds}\).
04

Finding Distance Traveled by Freeway Traffic

(c) Traffic on the freeway travels at a constant speed of 20 m/s. The time for the car to travel the ramp is 12 seconds, so the distance traveled by the traffic is calculated using \(d = v \times t\), where \(v = 20\, \text{m/s}\) and \(t = 12\, \text{s}\). Thus, \(d = 20 \times 12 = 240\, \text{m}\).

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Key Concepts

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

Constant Acceleration
Constant acceleration is a fascinating aspect of motion where an object's velocity changes at a steady rate over time. It implies that the acceleration remains the same whether the object is speeding up or slowing down. In this problem, the car on the ramp accelerates uniformly until it reaches a speed of 20 m/s. This scenario is ideal for applying kinematic equations, which effectively analyze such motion cases.
One significant characteristic of constant acceleration is the predictability it affords in calculations. Whether you're looking to determine how fast an object will be moving at a certain point or how far it will travel in a specific time, constant acceleration provides a straightforward pathway. As you work through the motion problem, recognizing the constant acceleration allows you to directly employ kinematic equations to answer questions about the car's travel along the ramp.
Kinematic Equations
Kinematic equations are powerful tools used to describe the motion of objects. They connect four essential variables: displacement (\( s \)), initial velocity (\( u \)), final velocity (\( v \)), and time (\( t \))—alongside acceleration (\( a \)). These equations assume constant acceleration and help solve various motion problems.
In our scenario, to determine the car's acceleration as it reaches the end of the ramp, utilize the equation: \( v^2 = u^2 + 2as \). Starting from rest, the initial velocity is zero (\( u = 0 \)). With the ramp 120 meters long, you can easily solve for acceleration (\( a \)).
By consistently applying these equations, students find themselves more comfortable dissecting motion problems. Each equation is a puzzle piece contributing to the overall image of an object's journey through time and space.
Motion Analysis
Motion analysis is the art of breaking down and understanding how objects move. It involves examining the path taken, the speed achieved, and the forces involved. In this problem, the car’s journey from a rest state to entering the freeway is broken into clear stages with the help of kinematic equations.
By analyzing the motion, you can uncover important insights such as how quickly the car accelerates and how quickly it reaches the required speed of freeway traffic. Understanding motion analysis enables you to appreciate how different elements of kinematics interlink to provide a cohesive understanding of the car’s journey on the ramp.
Moreover, this analysis helps solidify the fundamental concept that acceleration significantly impacts velocity and the time it takes for traveling distances. These skills are crucial not only in solving homework problems but also in appreciating real-world dynamics.
Velocity and Acceleration
Velocity, the speed at which an object moves in a particular direction, and acceleration, the rate at which velocity changes, are core concepts in kinematics. Understanding the relationship between these two is vital for analyzing motion effectively.
For the car accelerating up the ramp, velocity changes from zero to 20 m/s, illustrating acceleration's impact over time. Knowing both initial and final velocity states allows for the application of kinematic equations to determine the exact acceleration and the consequent time needed to reach the freeway speed.
The interplay between velocity and acceleration isn't just mathematical. It describes how objects behave in the real world, from the gradual pickup of speed in a car to the rapid deceleration when braking. Recognizing this relationship helps students predict and control motion across various contexts beyond theoretical exercises.

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Most popular questions from this chapter

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Determined to test the law of gravity for himself, a student walks off a skyscraper 180 \(\mathrm{m}\) high, stopwatch in hand, and starts his free fall (zero initial velocity). Five seconds later, Superman arrives at the scene and dives off the roof to save the student. Superman leaves the roof with an initial speed \(v_{0}\) that he produces by pushing himself downward from the edge of the roof with his legs of steel. He then falls with the same acceleration as any freely falling body. (a) What must the value of \(v_{0}\) be so that Superman catches the student just before they reach the ground? (b) On the same graph, sketch the positions of the student and of Superman as functions of time. Take Superman's initial speed to have the value calculated in part (a). (c) If the height of the skyscraper is less than some minimum value, even Superman can't reach the student before he hits the ground. What is this minimum height?

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