Question

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)?

Answer 1

Answer:

Explanation:

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