A light-rail train going from one station to the next on a straight section of track accelerates from rest at 1.1 {\rm m/s}^2 for 20 {\rm s}. It then proceeds at constant speed for 1100 {\rm m} before slowing down at 2.2 {\rm m/s}^2 until it stops at the station.
What is the distance between stations?
How much time does it take the train to go between stations?
It takes 80 seconds to travel between stations
How did you get 80 seconds?
To find the distance between stations, we need to determine the total distance traveled by the train during the acceleration and deceleration phases.
First, let's find the distance traveled during the acceleration phase.
The acceleration during this phase is 1.1 m/s^2, and the duration is 20 s.
We can use the kinematic equation: s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.
Since the train starts from rest (u = 0), the equation simplifies to s = (1/2)at^2.
Substituting the values, s = (1/2)(1.1)(20)^2 = 220 m.
Next, let's find the distance traveled at constant speed.
The train moves at a constant speed, so there is no acceleration. Therefore, the distance is simply the speed multiplied by the time.
Since the speed is constant, we can use the formula s = vt, where v is the speed and t is the time.
Substituting the values, s = (v)(1100).
Since the speed is not given directly, we need to find it. We know that speed is the rate of change of distance with respect to time. During the acceleration phase, the speed increases at a constant rate, so the average speed is half the final speed. Therefore, the final speed is 2 times the average speed.
The average speed during the acceleration phase can be found using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Since the train starts from rest (u = 0) and the acceleration is 1.1 m/s^2, the average speed can be calculated as v = (1/2)(1.1)(20) = 11 m/s.
Therefore, the final speed is 2 times the average speed, which is 22 m/s.
Substituting the values, s = (22)(1100) = 24200 m.
Finally, let's find the distance traveled during the deceleration phase.
The acceleration during this phase is -2.2 m/s^2 (negative because it is in the opposite direction of motion), and the train stops at the station (final velocity is 0).
Using the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance,
we can find s = (v^2 - u^2) / (2a).
Substituting the values, s = (0 - (22)^2) / (2(-2.2)) = 110 m.
Now we can calculate the total distance between stations by adding up the distances:
Total distance = 220 m + 1100 m + 110 m = 1430 m.
To find the total time taken by the train to go between stations, we need to add up the times taken during each phase.
The time taken during the acceleration phase is given as 20 s.
The time taken during the constant speed phase is equal to the distance divided by the speed.
Using the formula t = s / v, where t is the time, s is the distance, and v is the speed,
t = 1100 m / 22 m/s = 50 s.
The time taken during the deceleration phase is given as the time taken during the acceleration phase because the acceleration is the same in magnitude and duration.
Total time = 20 s + 50 s + 20 s = 90 s.
Therefore, the distance between stations is 1430 m, and it takes the train 90 seconds to go between stations.
1st acceleration: s = .5 * 1.1 * 400 = 220m
velocity attained: 1.1*20 = 22m/s
2nd acceleration: 22m/s / 2.2m/s^2 = 10s
2nd distance = .5 * 2.2 * 100 = 55m
distance = 220 + 1100 + 55 = 1375m