/*! 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 8 A Honda Civic travels in a strai... [FREE SOLUTION] | 91Ó°ÊÓ

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A Honda Civic travels in a straight line along a road. Its distance \(x\) from a stop sign is given as a function of time \(t\) by the equation \(x(t)=\alpha t^{2}-\beta t^{3},\) where \(\alpha=1.50 \mathrm{m} / \mathrm{s}^{2}\) and \(\beta=\) 0.0500 \(\mathrm{m} / \mathrm{s}^{3} .\) Calculate the average velocity of the car for each time interval: \((a) t=0\) to \(t=2.00 \mathrm{s} ;(b) t=0\) to \(t=4.00 \mathrm{s}\) (c) \(t=2.00 \mathrm{s}\) to \(t=4.00 \mathrm{s}\)

Short Answer

Expert verified
(a) 2.8 m/s, (b) 5.2 m/s, (c) 7.6 m/s.

Step by step solution

01

Understanding the Problem

We are given a function for the distance from a stop sign, \(x(t) = \alpha t^2 - \beta t^3\), where \(\alpha = 1.50 \text{ m/s}^2\) and \(\beta = 0.0500 \text{ m/s}^3\). We need to find the average velocity over specified time intervals.
02

Formula for Average Velocity

The average velocity \(v_{avg}\) over a time interval \([t_1, t_2]\) is given by \[v_{avg} = \frac{x(t_2) - x(t_1)}{t_2 - t_1}\]. We will use this formula to find the average velocities for different intervals.
03

Calculate Distance at Specific Times

First, calculate \(x(t)\) for the given times.- For \(t = 0\), \(x(0) = 1.50 \times 0^2 - 0.0500 \times 0^3 = 0\).- For \(t = 2\), \(x(2) = 1.50 \times 2^2 - 0.0500 \times 2^3 = 6 - 0.4 = 5.6\) meters.- For \(t = 4\), \(x(4) = 1.50 \times 4^2 - 0.0500 \times 4^3 = 24 - 3.2 = 20.8\) meters.
04

Calculate Average Velocity for Each Interval

Now calculate the average velocity for each interval using the formula from Step 2.(a) For \(t = 0\) to \(t = 2\):\[v_{avg} = \frac{x(2) - x(0)}{2 - 0} = \frac{5.6 - 0}{2} = 2.8 \text{ m/s}\](b) For \(t = 0\) to \(t = 4\):\[v_{avg} = \frac{x(4) - x(0)}{4 - 0} = \frac{20.8 - 0}{4} = 5.2 \text{ m/s}\](c) For \(t = 2\) to \(t = 4\):\[v_{avg} = \frac{x(4) - x(2)}{4 - 2} = \frac{20.8 - 5.6}{2} = 7.6 \text{ m/s}\]

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Key Concepts

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

Kinematics Equations
Understanding kinematics equations is crucial when studying motion, particularly in physics, where they describe the relationships between displacement, velocity, acceleration, and time. These equations make it easier to predict the future states of moving objects based on their current motions.
In this exercise, our kinematics equation is given by the distance function: \(x(t) = \alpha t^2 - \beta t^3\), where \(\alpha\) and \(\beta\) are constants. Such an equation is a specific example of how displacement can be calculated using time-dependent expressions in physics.
Key points to remember about kinematics equations:
  • They describe the motion of objects without considering the object's mass or forces acting on them.
  • The coefficients, like \(\alpha = 1.50 \, \text{m/s}^2\) and \(\beta = 0.0500 \, \text{m/s}^3\), provide specific motion characteristics like acceleration rates.
  • They help in finding other motion-related quantities like velocity and acceleration by differentiating or integrating the function.

These equations are vital for conducting experiments and solving problems related to linear motion, such as predicting how far a car travels over time.
Motion Along a Straight Line
Motion along a straight line is one of the simplest forms of motion and is often studied first in physics due to its straightforward nature. In this case, our Honda Civic car is traveling in a straight line, much like how many basic physics problems are set up to simplify calculations.
Some insights about linear motion:
  • In straight-line motion, only one spatial dimension is considered, often along an "x" axis, making it easier to calculate distances and velocities.
  • This motion can be uniform, with constant speed, or non-uniform, with changing speed and direction defined by a function of time like \(x(t) = \alpha t^2 - \beta t^3\).
  • Understanding an object’s motion in one dimension is foundational for grasping more complex motions in two or three dimensions.

When analyzing motion like this, scientists and engineers can predict future positions and times effectively, which is invaluable in areas such as transportation and safety engineering.
Calculating Displacement
Displacement refers to the change in position of an object, and it is a vector quantity, meaning it has both magnitude and direction. In this context, we're interested in how far the car travels from one point to another within specific time intervals.
Here's how displacement calculation works:
  • Use the given function, \( x(t) = \alpha t^2 - \beta t^3 \), to calculate positions at given times. For instance, \( x(2) \) and \( x(4) \) were found using the equation.
  • Displacement is the difference in these position values: \( x(t_2) - x(t_1) \).
  • The practice of finding position helps in determining other dynamics like speed or velocity, specifically over intervals, which results in average velocity calculations.

By understanding displacement and how to calculate it, students can better comprehend how objects move over time, providing the foundation for advanced study in fields like mechanics and dynamics.

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Most popular questions from this chapter

An automobile and a truck start from rest at the same instant, with the automobile initially at some distance behind the truck. The truck has a constant acceleration of \(2.10 \mathrm{m} / \mathrm{s}^{2},\) and the antomobile an acceleration of 3.40 \(\mathrm{m} / \mathrm{s}^{2} .\) The automobile overtakes the truck after the truck has moved 40.0 \(\mathrm{m}\) . (a) How much time does it take the automobile to overtake the truck? (b) How far was the automobile behind the truck initially? (c) What is the speed of each when they are abreast? (d) On a single graph, sketch the position of each vehicle as a function of time. Take \(x=0\) at the initial location of the truck.

At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 3.20 \(\mathrm{m} / \mathrm{s}^{2} .\) At the same instant a truck, traveling with a constant speed of 20.0 \(\mathrm{m} / \mathrm{s}\) , overtakes and passes the car. (a) How far beyond its starting point does the car overtake the truck? (b) How fast is the car traveling when it overtakes the truck? (c) Sketch an \(x\) -t graph of the motion of both vehicles. Take \(x=0\) at the inter-section. \(\left(\text { d) Sketch a } v_{x}-t \text { graph of the motion of both vehicles. }\right.\)

Two sunt drivers drive directly toward each other. At time \(t=0\) the two cars are a distance \(D\) apart, car 1 is at rest, and \(\operatorname{car} 2\) is moving to the left with speed \(v_{0}\) . Car 1 begins to move at \(t=0\) , speeding up with a constant acceleration \(a_{x}\) Car 2 continues to move with a constant velocity. (a) At what time do the two cars collide? \((b)\) Find the speed of car 1 just before it collides with \(\operatorname{car} 2 .\) (c) Sketch \(x-t\) and \(v_{x}-t\) graphs for car 1 and \(\operatorname{car} 2 .\) For each of the two graphs, draw the curves for both cars on the same set of axes.

Trip Home. You normally drive on the freeway between San Diego and Los Angeles at an average speed of 105 \(\mathrm{km} / \mathrm{h}\) \((65 \mathrm{mi} / \mathrm{h}),\) and the trip takes 2 \(\mathrm{h}\) and 20 \(\mathrm{min}\) . On a Friday afternoon, however, heavy traffic slows you down and you drive the same distance at an average speed of only 70 \(\mathrm{km} / \mathrm{h}\) (43 milh). How much longer does the trip take?

A certain volcano on earth can eject rocks vertically to a maximum height \(H .\) (a) How high (in terms of \(H\) ) would these rocks go if a volcano on Mars ejected them with the same initial velocity? The acceleration due to gravity on Mars is 3.71 \(\mathrm{m} / \mathrm{s}^{2}\) , and you can neglect air resistance on both planets. (b) If the rocks are in the air for a time \(T\) on earth, for how long (in terms of \(T )\) will they be in the air on Mars?

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