Question

A light-rail train going from one station to the next on a straight section of track accelerates from rest at 1.1 m/s^2 for 20s. It then proceeds at constant speed for 1100m before slowing down at 2.2m/s^2 until it stops at the station. A) What is the distance between stations? B) How much time does it take the train to go between the stations?

Answer 1

A) The distance between the stations is 1430m

B) The time it takes the train to go between the stations is 80s

Step-by-step explanation:

First we will calculate the distance covered for the first 20s.

From one the equations of kinematics for linear motion

`S = ut + (1)/(2)at^(2) \\`

Where
`S` is distance traveled

`u` is the initial velocity

`t` is time

and
`a` is acceleration

Since the train starts from rest,
`u` = 0 m/s

Hence, for the first 20s

`a =` 1.1 m/s²;
`t =` 20s,
`u` = 0 m/s

∴
`S = ut + (1)/(2)at^(2) \\` gives

`S = (0)(20) + (1)/(2)(1.1)(20)^(2)`

`S = (1)/(2)(1.1)(20)^(2)`

`S =` 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`

Where
`v` is the final velocity

Hence,

`v = 0 + 1.1(20)`

`v = 1.1(20)`

`v =` 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`

The initial velocity,
`u` here will be the final velocity before it started slowing down

∴
`u` = 22 m/s

The final velocity will be 0, since it came to a stop.

∴
`v` = 0 m/s

`a = -`2.2 m/s² ( - indicates deceleration)

Hence,

`v^(2) = u^(2) + 2as` gives

`0^(2) =22^(2) +2(-2.2)s`

`0=22^(2) - (4.4)s\\4.4s = 484\\s = (484)/(4.4) \\s = 110m`

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 =(Distance)/(Time)\\`

Then,

`Time = (Distance)/(Speed)`

`Time = (1100)/(22)`

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 = (22)/(2.2) \\t= 10 s`

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