/*! 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

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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

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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 \(d\).

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