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?
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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
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