/*! 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 35 A train is traveling south at \(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 35

A train is traveling south at \(24.0 \mathrm{m} / \mathrm{s}\) when the brakes are applied. It slows down with a constant acceleration to a speed of $6.00 \mathrm{m} / \mathrm{s}\( in a time of \)9.00 \mathrm{s} .$ (a) Draw a graph of \(v_{x}\) versus \(t\) for a 12 -s interval (starting \(2 \mathrm{s}\) before the brakes are applied and ending 1 s after the brakes are released). Let the \(x\) -axis point to the north. (b) What is the acceleration of the train during the \(9.00-\mathrm{s}\) interval? \((\mathrm{c})\) How far does the train travel during the $9.00 \mathrm{s} ?$

Short Answer

Expert verified
Question: A train traveling south slows down with constant acceleration from an initial velocity of 24.0 m/s to a final velocity of 6.00 m/s during a time interval of 9.00 s. (a) Draw a graph of \(v_x\) versus \(t\). (b) Calculate the acceleration of the train. (c) Determine the distance traveled by the train during this time interval when it is slowing down. Answer: (a) The graph of \(v_x\) versus \(t\) will be a straight line with a negative slope, connecting the points (-24.0 m/s, 0 s) and (-6.00 m/s, 9.00 s). (b) The acceleration of the train is -2.00 m/s虏. (c) The distance traveled by the train during the deceleration process is 135 m.

Step by step solution

01

Draw a graph of \(v_x\) versus \(t\)

To draw a graph of \(v_x\) versus \(t\), we need to find the values of the initial and final velocities on the x-axis. Since the initial velocity is in the south direction, which is the negative x-direction, the initial velocity is \(-24.0\, \mathrm{m/s}\). Similarly, the final velocity will be \(-6.00\, \mathrm{m/s}\). Plot these values on the graph, along with the time interval from \(-2\, \mathrm{s}\) to \(12\, \mathrm{s}\). Draw the straight line connecting all the points for the time interval starting \(2\, \mathrm{s}\) before the brakes are applied and ending \(1\, \mathrm{s}\) after the brakes are released.
02

Calculate the acceleration of the train

To find the acceleration, we need to use the formula: \(a = \dfrac{v_f - v_i}{t}\) where \(v_i\) is the initial velocity, \(v_f\) is the final velocity, and \(t\) is the time interval. With the given values, we have: \(a = \dfrac{-6.00\, \mathrm{m/s} - (-24.0\, \mathrm{m/s})}{9.00\, \mathrm{s}}\) Calculate the value of \(a\).
03

Calculate the distance traveled by the train during the deceleration process

To find the distance traveled, we can use the formula: \(d = v_i t + \dfrac{1}{2}at^2\) Using the obtained value for \(a\) (from step 2), the given values for \(v_i\), and \(t, \) plug in the values into the formula and calculate the distance \(d\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In the problems, please assume the free-fall acceleration $g=9.80 \mathrm{m} / \mathrm{s}^{2}$ unless a more precise value is given in the problem statement. Ignore air resistance. A rocket is launched from rest. After 8.0 min, it is \(160 \mathrm{km}\) above the Earth's surface and is moving at a speed of $7.6 \mathrm{km} / \mathrm{s} .$ Assuming the rocket moves up in a straight line, what are its (a) average velocity and (b) average acceleration?
In the human nervous system, signals are transmitted along neurons as action potentials that travel at speeds of up to \(100 \mathrm{m} / \mathrm{s} .\) (An action potential is a traveling influx of sodium ions through the membrane of a neuron.) The signal is passed from one neuron to another by the release of neurotransmitters in the synapse. Suppose someone steps on your toe. The pain signal travels along a 1.0 -m-long sensory neuron to the spinal column, across a synapse to a second 1.0 -m-long neuron, and across a second synapse to the brain. Suppose that the synapses are each \(100 \mathrm{nm}\) wide, that it takes \(0.10 \mathrm{ms}\) for the signal to cross each synapse, and that the action potentials travel at \(100 \mathrm{m} / \mathrm{s} .\) (a) At what average speed does the signal cross a synapse? (b) How long does it take the signal to reach the brain? (c) What is the average speed of propagation of the signal?
please assume the free-fall acceleration \(g=9.80 \mathrm{m} / \mathrm{s}^{2}\) unless a more precise value is given in the problem statement. Ignore air resistance. A 55 -kg lead ball is dropped from the leaning tower of Pisa. The tower is 55 m high. (a) How far does the ball fall in the first 3.0 s of flight? (b) What is the speed of the ball after it has traveled 2.5 m downward? (c) What is the speed of the ball 3.0 s after it is released? (d) If the ball is thrown vertically upward from the top of the tower with an initial speed of $4.80 \mathrm{m} / \mathrm{s}$, where will it be after 2.42 s?
please assume the free-fall acceleration \(g=9.80 \mathrm{m} / \mathrm{s}^{2}\) unless a more precise value is given in the problem statement. Ignore air resistance. A penny is dropped from the observation deck of the Empire State building (369 m above ground). With what velocity does it strike the ground?
A car is speeding up and has an instantaneous velocity of $1.0 \mathrm{m} / \mathrm{s}\( in the \)+x\( -direction when a stopwatch reads \)10.0 \mathrm{s} .$ It has a constant acceleration of \(2.0 \mathrm{m} / \mathrm{s}^{2}\) in the \(+x\) -direction. (a) What change in speed occurs between \(t=10.0 \mathrm{s}\) and \(t=12.0 \mathrm{s} ?(\mathrm{b})\) What is the speed when the stopwatch reads \(12.0 \mathrm{s} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Electrostatics

Read Explanation

Electromagnetism

Read Explanation

Medical Physics

Read Explanation

Classical Mechanics

Read Explanation

Electric Charge Field and Potential

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.