A train backs up from an initial velocity of -4.0 m/s and an acceleration of -0.27 m/s^2. What is the train's velocity after 17 s?

the final velocity of the train is -4.99 m/s.

Explanation:

Given;

initial velocity of the train, u = -0.4 m/s

acceleration of the train, a = -0.27 m/s²

time of motion of the train, t = 17 s

The final velocity of the train is calculated as follows;

v = u + at

where;

v is the final velocity of the train

v = -0.4 + (-0.27 x 17)

v = -0.4 - 4.59

v = -4.99 m/s

Therefore, the final velocity of the train is -4.99 m/s.

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow