Respuesta :
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
[tex]S = ut + \frac{1}{2}at^{2} \\[/tex]
Where [tex]S[/tex] is distance traveled
[tex]u[/tex] is the initial velocity
[tex]t[/tex] is time
and [tex]a[/tex] is acceleration
Since the train starts from rest, [tex]u[/tex] = 0 m/s
Hence, for the first 20s
[tex]a =[/tex] 1.1 m/s²; [tex]t =[/tex] 20s, [tex]u[/tex] = 0 m/s
∴ [tex]S = ut + \frac{1}{2}at^{2} \\[/tex] gives
[tex]S = (0)(20) + \frac{1}{2}(1.1)(20)^{2}[/tex]
[tex]S = \frac{1}{2}(1.1)(20)^{2}[/tex]
[tex]S =[/tex] 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
[tex]v = u +at[/tex]
Where [tex]v[/tex] is the final velocity
Hence,
[tex]v = 0 + 1.1(20)[/tex]
[tex]v = 1.1(20)[/tex]
[tex]v =[/tex] 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
[tex]v^{2} = u^{2} + 2as[/tex]
The initial velocity, [tex]u[/tex] here will be the final velocity before it started slowing down
∴[tex]u[/tex] = 22 m/s
The final velocity will be 0, since it came to a stop.
∴ [tex]v[/tex] = 0 m/s
[tex]a = -[/tex]2.2 m/s² ( - indicates deceleration)
Hence,
[tex]v^{2} = u^{2} + 2as[/tex] gives
[tex]0^{2} =22^{2} +2(-2.2)s[/tex]
[tex]0=22^{2} - (4.4)s\\4.4s = 484\\s = \frac{484}{4.4} \\s = 110m[/tex]
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,
[tex]Speed =\frac{Distance}{Time}\\[/tex]
Then,
[tex]Time = \frac{Distance}{Speed}[/tex]
[tex]Time = \frac{1100}{22}[/tex]
Time = 50 s
And then, the time spent while decelerating (that is, while slowing down)
From,
[tex]v = u + at\\0 = 22 +(-2.2)t\\2.2t = 22\\t = \frac{22}{2.2} \\t= 10 s[/tex]
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
