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 min, 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?

a) The final velocity is 28.0 m/s

b) It will take 50.9 s to come to a stop from a velocity of 28.0 m/s

c) In the first case ( point "a") the train will travel 7.68 × 10³ m. In the secid case, the train will travel 713 m

Explanation:

The equations for the position and velocity of objects moving in a straight line are as follows:

x = x0 + v0 • t + 1/2 • a •t²

v = v0 + a • t

Where

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

a) Using the equation of velocity:

v = v0 + a • t

v = 4.00 m/s + 0.0500 m/s² • 480 s

v = 28.0 m/s

b) Now, the initial velocity is 28.0 m/s and the final velocity is 0 m/s. Then:

0 m/s = 28.0 m/s - 0.550 m/s²• t

-28.0 m/s / - 0.550 m/s² = t

t = 50.9 s

c) Now, we have to use the equation for the position of the train.

x = x0 + v0 • t + 1/2 • a •t² (let´s place the center of the frame of reference at x0. In this way, x0 = 0)

In the first case:

x = 4.00 m/s • 480 s + 1/2 • 0.0500 m/s² • (480 s)²

x = 7.68 × 10³ m

In the second case:

x = 28.0 m/s • 50.9 s - 1/2 • 0.550 m/s² • (50.9 s)²

x = 713 m

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the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

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