. 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
1 answer:
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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