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. An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of 0.150 m/s2 as it goes through. The station is 210 m long. (a) How long is the nose of the train in the station

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Answer:

Explanation:

Give it that,

Initial velocity

u = 22m/s

Deceleration a = - 0.15m/s2

Time taken to travel a station long of 210m

Using equation of motion

Let know the final velocity, when it leaves the station

v² = u²+2as

v² = 22²+2×(-0.15)×210

v² = 484—63

v² = 421

v =√421

v = 20.52m/s

Then,

Using equation of motion to find time taken

v = u + at

20.52 = 22 +(-0.15)t

20.52-22 = -0.15t

-0.15t = -1.48

t = -1.48/-0.15

t = 9.88 sec

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