Problem:
Freight trains can produce only relatively small accelerations and decelerations.
(a) What is the final velocity of a freight train that accelerates at a rate of 0.0500 m/s2 for 8.00 minutes, starting with an initial velocity of 4.00 m/s?
(b) If the train can slow down at a rate of 0.550 m/s2, how long will it take to come to a stop from this velocity?
(c) How far will it travel in each case?
Solution:
Part A
We are given the the following: a=0.0500 \ \text{m/s}^2; t=8.00 \ \text{mins}; and v_0=4.00 \ \text{m/s}.
The final velocity can be solved using the formula v_f=v_0+at. We substitute the given values.
\begin{align*}
v_f& = v_0+at \\
v_f & = 4.00\:\text{m/s}+\left(0.0500\:\text{m/s}^2\right)\left(8.00\:\text{min}\times \frac{60\:\sec }{1\:\min }\right) \\
v_f & = 28.0 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}Part B
Rearrange the equation we used in part (a) by solving in terms of t, we have
\begin{align*}
t & =\frac{{v_f}-v_0}{a} \\
t & = \frac{0\:\text{m/s}-28\:\text{m/s}}{-0.550\:\text{m/s}^2} \\
t & = 50.91\:\sec\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}Part C
The change in position for part (a), \Delta x, or distance traveled is computed using the formula \Delta x=v_0 t+\frac{1}{2} at^2.
\begin{align*}
\Delta x & =v_0 t+\frac{1}{2} at^2 \\
\Delta x & =\left(4.0\:\text{m/s}\right)\left(480\:\text{s}\right)+\frac{1}{2}\left(0.0500\:\text{m/s}^2\right)\left(480\:\text{s}\right)^2 \\
\Delta x & = 7680\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}For the situation in part (b), the distance traveled is computed using the formula \Delta x=\frac{v_f^2-v_0^2}{2 a}.
\begin{align*}
\Delta x & =\frac{\left(0\:\text{m/s}\right)^2-\left(28.0\:\text{m/s}\right)^2}{2\left(-0.550\:\text{m/s}^2\right)} \\
\Delta x & =712.73\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}College Physics Chapter 2 Problems

College Physics 2nd Edition Solutions Table of Contents
Chapter 1: Introduction: The Nature of Science and Physics
Chapter 3: Two-Dimensional Kinematics
Chapter 4: Dynamics: Force and Newton’s Law of Motion
Chapter 5: Further Applications of Newton’s Laws: Friction, Drag, and Elasticity
Chapter 6: Uniform Circular Motion and Gravitation
Chapter 7: Work, Energy, and Energy Resources
Chapter 8: Linear Momentum and Collisions
Chapter 10: Rotational Motion and Angular Momentum
Chapter 12: Fluid Dynamics and Its Biological and Medical Applications
Chapter 13: Temperature, Kinetic Theory, and the Gas Laws
Chapter 14: Heat and Heat Transfer Methods
Chapter 15: Thermodynamics
Chapter 16: Oscillatory Motion and Waves
Chapter 17: Physics of Hearing
Chapter 18: Electric Charge and Electric Field
Chapter 19: Electric Potential and Electric Field
Chapter 20:
Electric Current, Resistance, and Ohm’s Law
Chapter 21: Circuits and DC Instruments
Chapter 22: Magnetism
Chapter 23: Electromagnetic Induction, AC Circuits, and Electrical Technologies
Chapter 24: Electromagnetic Waves
Chapter 25: Geometric Optics
Chapter 26: Vision and Optical Instrument
Chapter 27: Wave Optics
Chapter 28: Special Relativity
Chapter 29: Introduction to Quantum Physics
Chapter 30: Atomic Physics
Chapter 31: Radioactivity and Nuclear Physics
Chapter 32: Medical Applications of Nuclear Physics
Chapter 33: Particle Physics
Chapter 34: Frontiers of Physics
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