Freight trains can produce only relatively small accelerations and decelerations. What is the final velocity, in meters per second, of a freight train that accelerates at a rate of 0.065 m/s^2 for 9.75 min, starting with an initial velocity of 3.4 m/s? If the train can slow down at a rate of 0.625 m/s^2, how long, in seconds, does it take to come to a stop from this velocity? How far, in meters, does the train travel during the process described in part (a) ? How far, in meters, does the train travel during the process described in part (b) ?

Question

Genevieve

Physics

31 August, 01:55

Freight trains can produce only relatively small accelerations and decelerations. What is the final velocity, in meters per second, of a freight train that accelerates at a rate of 0.065 m/s^2 for 9.75 min, starting with an initial velocity of 3.4 m/s? If the train can slow down at a rate of 0.625 m/s^2, how long, in seconds, does it take to come to a stop from this velocity? How far, in meters, does the train travel during the process described in part (a) ? How far, in meters, does the train travel during the process described in part (b) ?

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Answers (1)

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Mara Yang · Answer 1

Initial velocity of train u = 3.4 m⁻¹, Acceleration a =.065 ms⁻²,

time t = 9.75 x 60 = 585 s.

v = u + at

= 3.4 +.065 x 585

= 41.425 m / s

distance travelled during the acceleration (s)

s = ut + 1/2 at²

= 3.4 x 585 +.5 x. 065 x 585²

= 1989 + 11122.31

= 13111.31 m.

again for slowing process

u = 41.425 m / s

v = u - at

0 = 41.425 - 0.625 t

t = 66.28 s

v² = u² - 2as

0 = 41.425² - 2 x. 625 s

s = 1372.82 m