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