Question

A train moving with a velocity of 42.9 km/hour North, increases its speed with a uniform acceleration of 0.250 m/s2 North until it reaches a velocity of 160.0 km/hour North. What distance did the train travel while it was increasing its velocity, in units of meters?

Answer 1

Answer:

3672.5m

Explanation:

V=u+at

v---> final velocity =160km/hour =44.5m/s

u---->initial velocity =42.9km/hour=12m/s

a--->acceleration =0.25m/s^2

therefore time t is 130s

v^2-u^2=2as

therefore distance s =

{v^(2)-u^(2)}/2a

by sub the values,

we get s=3672.5m