Question

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?

Answer 1

Answer:

v = 43.72 m/s

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,

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

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