Answer:
A) The distance between the stations is 1430m
B) The time it takes the train to go between the stations is 80s
Explanation:
First we will calculate the distance covered for the first 20s.
From one the equations of kinematics for linear motion
![S = ut + \frac{1}{2}at^{2} \\](https://tex.z-dn.net/?f=S%20%3D%20ut%20%2B%20%5Cfrac%7B1%7D%7B2%7Dat%5E%7B2%7D%20%20%5C%5C)
Where
is distance traveled
is the initial velocity
is time
and
is acceleration
Since the train starts from rest,
= 0 m/s
Hence, for the first 20s
1.1 m/s²;
20s,
= 0 m/s
∴
gives
![S = (0)(20) + \frac{1}{2}(1.1)(20)^{2}](https://tex.z-dn.net/?f=S%20%3D%20%280%29%2820%29%20%2B%20%5Cfrac%7B1%7D%7B2%7D%281.1%29%2820%29%5E%7B2%7D)
![S = \frac{1}{2}(1.1)(20)^{2}](https://tex.z-dn.net/?f=S%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%281.1%29%2820%29%5E%7B2%7D)
220m
This is the distance covered in the first 20s.
- The train then proceeds at constant speed for 1100m.
Now, we will calculate the speed attained here
From
![v = u +at](https://tex.z-dn.net/?f=v%20%3D%20u%20%2Bat)
Where
is the final velocity
Hence,
![v = 0 + 1.1(20)](https://tex.z-dn.net/?f=v%20%3D%200%20%2B%201.1%2820%29)
![v = 1.1(20)](https://tex.z-dn.net/?f=v%20%3D%201.1%2820%29)
22 m/s
This is the constant speed attained when it proceeds for 1100m
- The train then slows down at a rate of 2.2 m/s² until it stops
We can calculate the distance covered while slowing down from
![v^{2} = u^{2} + 2as](https://tex.z-dn.net/?f=v%5E%7B2%7D%20%3D%20u%5E%7B2%7D%20%2B%202as)
The initial velocity,
here will be the final velocity before it started slowing down
∴
= 22 m/s
The final velocity will be 0, since it came to a stop.
∴
= 0 m/s
2.2 m/s² ( - indicates deceleration)
Hence,
gives
![0^{2} =22^{2} +2(-2.2)s](https://tex.z-dn.net/?f=0%5E%7B2%7D%20%3D22%5E%7B2%7D%20%2B2%28-2.2%29s)
![0=22^{2} - (4.4)s\\4.4s = 484\\s = \frac{484}{4.4} \\s = 110m](https://tex.z-dn.net/?f=0%3D22%5E%7B2%7D%20-%20%284.4%29s%5C%5C4.4s%20%3D%20484%5C%5Cs%20%3D%20%5Cfrac%7B484%7D%7B4.4%7D%20%5C%5Cs%20%3D%20110m)
This is the distance traveled while slowing down.
A) The distance between the stations is
220m + 1100m + 110m
= 1430m
Hence, the distance between the stations is 1430m
B) The time it takes the train to go between the stations
The time spent while accelerating at 1.1 m/s² is 20s
We will calculate the time spent when it proceeds at a constant speed of 22 m/s for 1100m,
From,
![Speed =\frac{Distance}{Time}\\](https://tex.z-dn.net/?f=Speed%20%3D%5Cfrac%7BDistance%7D%7BTime%7D%5C%5C)
Then,
![Time = \frac{Distance}{Speed}](https://tex.z-dn.net/?f=Time%20%3D%20%5Cfrac%7BDistance%7D%7BSpeed%7D)
![Time = \frac{1100}{22}](https://tex.z-dn.net/?f=Time%20%3D%20%5Cfrac%7B1100%7D%7B22%7D)
Time = 50 s
And then, the time spent while decelerating (that is, while slowing down)
From,
![v = u + at\\0 = 22 +(-2.2)t\\2.2t = 22\\t = \frac{22}{2.2} \\t= 10 s](https://tex.z-dn.net/?f=v%20%3D%20u%20%2B%20at%5C%5C0%20%3D%2022%20%2B%28-2.2%29t%5C%5C2.2t%20%3D%2022%5C%5Ct%20%3D%20%5Cfrac%7B22%7D%7B2.2%7D%20%5C%5Ct%3D%2010%20s)
This is the time spent while slowing down until it stops at the station.
Hence, The time it takes the train to go between the stations is
20s + 50s + 10s = 80s
The time it takes the train to go between the stations is 80s