If the velocity of a train which starts from rest is 72km/h after 5minutes, find out it's acceleration and distance travelled by the train in this time.
Answers
Given data :
★ initial velocity of train, u = 0
★ final velocity of train, v = 72 km/hr
★ time taken by train, t = 5 minute
★ acceleration, a = ?
★ displacement, s = ?
Solution : Firstly we have to find convert 72 km/hr into m/s and 5 minute into second.
⟹ 72 km/hr = (72 * 1000)/3600 m/s
⟹ 72 km/hr = 72000/3600 m/s
⟹ 72 km/hr = 20 m/s
and
⟹ 5 minute = 5 * 60 sec
⟹ 5 minute = 300 sec
Now, by kinematical equation
⟹ v = u + at
⟹ v - u = at
⟹ 20 - 0 = 300/a
⟹ 20 = 300a
⟹ a = 20/300
⟹ a = 1/15 m/s²
Now,
⟹ s = ut + ½ at²
⟹ s = 0 * 300 + ½ * 1/15 * (300)²
⟹ s = 0 + 1/30 * 90000
⟹ s = 3000 m
Answer : Acceleration of the train is 1/15 m/s² and distance covered by the train is 3000 m.
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