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First off Thank you! Please don\'t just provide an answer if you can please expl

ID: 1651579 • Letter: F

Question

First off Thank you! Please don't just provide an answer if you can please explain which formula you used and show work. I have the answers Im really just trying to figure out how to get them without an answer key, thanks again.

Problem 5: An express train passes through a station that is 210 m long. It enters with an initial speed of 22 m/s and decelerates at a rate of 0.15 m/s2 as it goes through. Part (a) During what time interval, in seconds, is the front of the train within the station? Numeric :A numeric value is expected and not an expression t= Part (b) How fast is the train moving, in meters per second, when its front leaves the station? Numeric :A numeric value is expected and not an expression v= Part (c) If the train is 132 m long, how long, in seconds, after the front of the train enters the station does the end of the train leave the station? Numeric :A numeric value is expected and not an expression Part (d) What is the velocity of the end of the train as it leaves in m/s? Numeric : A numeric value is expected and not an expression v=

Explanation / Answer

a] Front of the train will be entirely inside when D = 210m

use v2 = u2 + 2aD

=> v = [222 - 2(0.15)(210)] 1/2 = 20.518 m/s

and v = u + at

=> 20.518 = 22 - 0.15t

=> t = 9.88 s

b] when its front leaves the station, the velocity of the train will be: v = 20.518 m/s

c] Additional distance = d = 132m

velocity of the train when it covers an additional 132 m will be:

V = [(20.518)2 - 2(0.15)(132)]1/2 = 19.53 m/s

for this velocity,

19.53 = 20.518 - 0.15t

=> t = 6.592s

therefore the time (since the front of the train enters the station) when the back leaves the station will be:

T = 9.88 + 6.592 = 16.472s.

d] Velocity of the end of the train as it leaves the station = V = 19.53 m/s.

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