Motion of a train between two stations:
When the train starts from the station A to station B on a non-stop journey, its speed can have patterns. When at the station A, it is at rest. Then it moves and gains velocity. As distance increases velocity also increases and at 1 point of time it reaches the maximum, which is maintained for the same distance that is the velocity remains constant and then velocity begins decreasing and becomes zero finally, when the train reaches to the station B.
Example: A body is having velocity of 15 m/sec at the certain instant and 10 seconds later it has the velocity 45 m/sec. If velocity changes uniformly, find space described.
Solution: We have,
u = initial velocity of particle = 15 m/sec
a= acceleration of particle
v = final velocity of particle = 45 m/sec
the particle takes 10 seconds to acquire velocity of 45 m/sec
so v = u + at => 45 = 15 + 10 a => a = 3 m/sec2
Let space described be s meter, then v2 = u2 + 2as
so, (45)2 = (15)2 + 2 x 3 => s => s = 300 meters.
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