Freight trains can produce only relatively small...

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 $\mathrm{m} / \mathrm{s}^{2}$ for 8.00 min, starting with an initial velocity of 4.00 $\mathrm{m} / \mathrm{s} ?$ (b) If the train can slow down at a rate of 0.550 $\mathrm{m} / \mathrm{s}^{2}$ , how long will it take to come to a stop from this velocity? (c) How far will it travel in each case?

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Key Concepts

Constant Acceleration

This concept involves motion where the acceleration remains the same throughout the observed time period. In such situations, the standard kinematic equations, which relate velocity, displacement, acceleration, and time, can be applied. Whether the object is speeding up (accelerating) or slowing down (decelerating), the same formulas are used with appropriate signs for acceleration.

Kinematic Equations

These equations describe the motion of objects under constant acceleration. Key formulas include v = v? + at (which relates initial velocity, acceleration, time, and final velocity) and s = v?t + ½ at² (which gives the displacement traveled over a time period). They are essential tools for solving problems involving changes in velocity and calculating the distance covered during acceleration or deceleration.

Time Conversion

Time conversion is the process of converting units of time to ensure consistency in equations. For example, converting minutes to seconds is crucial so that the acceleration in meters per second squared properly combines with time in seconds in the kinematic equations.

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