/*! 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 67 A car and a truck are heading di... [FREE SOLUTION] | 91Ó°ÊÓ

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A car and a truck are heading directly toward one another on a straight and narrow street, but they avoid a head-on collision by simultaneously applying their brakes at \(t=0\). The resulting velocity-versus-time graphs are shown in Figure \(2-31\). What is the separation between the car and the truck when they have come to rest, given that at \(t=0\) the car is at \(x=15 \mathrm{m}\) and the truck is at \(x=-35 \mathrm{m} ?\) (Note that this information determines which line in the graph corresponds to which vehicle.)

Short Answer

Expert verified
The separation is the initial distance minus the combined displacements of the car and truck.

Step by step solution

01

Understand the Problem

We have two vehicles: a car and a truck that both start braking at the same time to avoid a collision. We need to find the distance between them when they have both stopped. Initially, the car is at position \( x = 15 \, \mathrm{m} \) and the truck is at position \( x = -35 \, \mathrm{m} \).
02

Identify Key Variables

We are provided initial positions and we know both vehicles decelerate to a stop. We will use the velocity-time graphs to determine the distance each vehicle travels until it stops.
03

Analyze the Velocity-Time Graphs

If the graphs show the velocity over time for both vehicles, the area under the velocity-time graph represents the change in position or displacement. Calculate this area for both vehicles.
04

Calculate the Car's Displacement

Find the area under the car's velocity-time graph up to the point where its velocity reaches zero. This area will give the total distance the car travels while stopping.
05

Calculate the Truck's Displacement

Similarly, find the area under the truck's velocity-time graph until it stops. The area represents the total distance the truck travels while decelerating.
06

Determine the Final Positions

Using the initial positions and the calculated displacements, find the final positions of both the car and the truck. \( \text{Final position of car} = \text{initial position} + \text{car's displacement} \) and \( \text{Final position of truck} = \text{initial position} + \text{truck's displacement} \).
07

Calculate the Final Separation

The separation between the car and the truck is the absolute difference between their final positions. Therefore, the separation is \( \lvert \text{Final position of car} - \text{Final position of truck} \rvert \).

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Key Concepts

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

Understanding Velocity-Time Graphs
Velocity-time graphs are a visual way to show how an object's velocity changes over time. The horizontal axis typically represents time, while the vertical axis represents velocity. One of the key aspects of these graphs is that the slope of the line at any given point represents the acceleration.
For a straight, sloped line, the object is experiencing constant acceleration or deceleration depending on the slope direction: positive for acceleration, negative for deceleration.
Importantly, the area under the curve of a velocity-time graph represents the total displacement: the distance and direction the object has traveled in that time frame.
  • If the graph is above the horizontal axis, the displacement occurs in the positive direction.
  • If it's below, it occurs in the negative direction.
This makes velocity-time graphs an essential tool for understanding motion dynamics, especially when evaluating braking scenarios for vehicles like in this exercise.
Calculating Displacement from Graphs
Displacement refers to the change in position of an object from a starting point. In the context of the problem, it gives us the distance each vehicle travels while decelerating until they stop. To find the displacement of each vehicle, look at the velocity-time graph for each.
The area under the graph line for each vehicle corresponds to its displacement during the braking period. This can be calculated using basic geometry:
  • For a simple triangular shape (common when vehicles decelerate uniformly), the area is calculated using \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \), where the base is the time, and the height is the initial velocity.
  • If the graph shape is more complex, break it down into simple geometric shapes like rectangles and triangles, then add their areas together for total displacement.
By determining the displacements, you can ascertain how far each vehicle travels while stopping.
Assessing Initial Positions
Initial positions are crucial for determining where the vehicles start from and understanding how their displacements affect their final positions. In this problem, initial positions help us identify which graph pertains to each vehicle.
At the start time \( t = 0 \), the car is positioned at 15 meters, and the truck is at \( -35 \) meters relative to a common reference point. These are not merely starting locations. They serve as baselines to assess how far each vehicle moves and where they each come to a stop.
  • The initial positions help us know from which point to measure the change in location due to the displacement during braking.
  • By calculating the displacement, we can add this to the initial position of each vehicle to find their final resting points."
Hence, initial positions offer a starting point to monitor changes in movement and position effectively.
Understanding Deceleration
Deceleration is the process of reducing speed or velocity, essentially negative acceleration. When the car and the truck apply brakes, they decelerate to a stop. In velocity-time graphs, deceleration is represented by a negative slope.
Deceleration is important because it influences how quickly a vehicle comes to rest. Factors that can affect deceleration include:
  • Vehicle mass
  • Brake efficiency
  • Road surface conditions
In a real-world situation, precise deceleration rates could vary significantly depending on these factors.
For initial analysis, understanding deceleration in a physics context involves looking at how the change in speed over time impacts the distance traveled (displacement) before stopping. Calculating the rate of deceleration can also offer insights into how quickly each vehicle reduces its velocity to zero, given by the slope of the velocity-time graph.

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Most popular questions from this chapter

The world's highest fountain of water is located, appropriately enough, in Fountain Hills, Arizona. The fountain rises to a height of \(560 \mathrm{ft}(5\) feet higher than the Washington Monument). (a) What is the initial speed of the water? (b) How long does it take for water to reach the top of the fountain?

You drive your car in a straight line at \(15 \mathrm{m} / \mathrm{s}\) for \(10 \mathrm{kilometers},\) then at \(25 \mathrm{m} / \mathrm{s}\) for another \(10 \mathrm{kilome}^{-}\) ters. (a) Is your average speed for the entire trip more than, less than, or equal to \(20 \mathrm{m} / \mathrm{s} ?\) (b) Choose the best explanation from among the following: I. More time is spent at \(15 \mathrm{m} / \mathrm{s}\) than at \(25 \mathrm{m} / \mathrm{s}\). II. The average of \(15 \mathrm{m} / \mathrm{s}\) and \(25 \mathrm{m} / \mathrm{s}\) is \(20 \mathrm{m} / \mathrm{s}\) III. Less time is spent at \(15 \mathrm{m} / \mathrm{s}\) than at \(25 \mathrm{m} / \mathrm{s}\).

A model rocket rises with constant acceleration to a height of \(3.2 \mathrm{m},\) at which point its speed is \(26.0 \mathrm{m} / \mathrm{s} .\) (a) How much time does it take for the rocket to reach this height? (b) What was the magnitude of the rocket's acceleration? (c) Find the height and speed of the rocket \(0.10 \mathrm{s}\) after launch.

A youngster bounces straight up and down on a trampoline. Suppose she doubles her initial speed from \(2.0 \mathrm{m} / \mathrm{s}\) to \(4.0 \mathrm{m} / \mathrm{s} .\) (a) By what factor does her time in the air increase? (b) By what factor does her maximum height increase? (c) Verify your answers to parts (a) and (b) with an explicit calculation.

To find the height of an overhead power line, you throw a ball straight upward. The ball passes the line on the way up after \(0.75 \mathrm{s},\) and passes it again on the way down \(1.5 \mathrm{s}\) after it was tossed. What are the height of the power line and the initial speed of the ball?

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