A Train is going with a speed of 29 m/s . if its acceleration is 4m/s . How much distance will it cover in 15 min. and what will be its velocity .
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Answer:
The distance covered by the train in 15 mins is 16,46,100 m and its velocity will be 3629 m/s.
Explanation:
The initial velocity of the train, u = 29 m/s
The acceleration of the train, a = 4 m/s
The time taken by the train, t = 15 min = 900 s
The distance covered by the train during this time interval can be calculated using the 2nd equation of motion.
S = ut + (at²/2)
S = (29 × 900) + (4 × 900² / 2)
S = 26100 + 1620000
S = 16,46,100 m
The final velocity of the train can be calculated using the 1st equation of motion.
v = u + at
v = 29 + (4 × 900)
v = 29 + 3600
v = 3629 m/s
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