16,213 views
21 votes
21 votes
A train with an initial velocity of 31 m/s begins accelerating at rate of 0.0705 m/s^2. If the train travels for 180.5s, how far does it travel?

User Mike Feltman
by
2.9k points

1 Answer

12 votes
12 votes

Answer:

v = 43.72 m/s

Step-by-step explanation:

Given that,

Initial velocity of the train, u = 31 m/s

Acceleration of the train, a = 0.0705 m/s²

Time for which the train travel, t = 180.5 s

We need to find the final velocity of the train. Let it is v.. Using first equation of kinematics to find it such that,


v=u+at\\\\v=31+0.0705* 180.5\\\\v=43.72\ m/s

So the final speed of the train is equal to 43.72m/s.

User Saurish Kar
by
3.3k points