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 20 m . It then proceeds at constant speed for 110m before slowing down at 2.2 m/s^2 until it stops at the station.
You forgot to ask a question. All you did was provide data.
1440m
To find the total time it takes for the light-rail train to travel from one station to the next, we can break down the motion into three parts: acceleration, constant speed, and deceleration.
1. Acceleration:
The train accelerates from rest at 1.1 m/s^2 for 20 m. To find the time it takes during this acceleration phase, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
- v is the final velocity (unknown)
- u is the initial velocity (0 m/s since it starts from rest)
- a is the acceleration (1.1 m/s^2)
- s is the displacement (20 m)
Rearranging the equation, we have:
v^2 = 0^2 + 2(1.1)(20)
v^2 = 44
v = √44
v ≈ 6.63 m/s
Using the equation v = u + at, we can find the time taken during acceleration:
6.63 = 0 + 1.1t
t = 6.63/1.1
t ≈ 6 seconds
So, it takes approximately 6 seconds for the train to accelerate.
2. Constant Speed:
The train proceeds at a constant speed for 110 m. Since the speed remains constant, the time taken during this phase can be determined using the formula:
time = distance/speed
Therefore, the time taken for this phase is:
time = 110 m / 6.63 m/s (using the speed calculated during acceleration)
time ≈ 16.59 seconds
So, it takes approximately 16.59 seconds for the train to travel at a constant speed.
3. Deceleration:
The train slows down at 2.2 m/s^2 until it stops at the station. Using the equation v^2 = u^2 + 2as, where v is the final velocity (0 m/s), u is the initial velocity (the constant speed), a is the deceleration, and s is the displacement (unknown), we can solve for s:
0^2 = (6.63)^2 + 2(-2.2)s
0 = 43.89 - 4.4s
4.4s = 43.89
s ≈ 9.97 m
To find the time taken during deceleration, we can use the equation v = u + at:
0 = 6.63 + (-2.2)t
-2.2t = -6.63
t ≈ 3 seconds
So, it takes approximately 3 seconds for the train to decelerate.
Finally, to find the total time taken for the entire journey, we add the times taken during each phase:
Total time = Acceleration time + Constant speed time + Deceleration time
Total time ≈ 6 seconds + 16.59 seconds + 3 seconds
Total time ≈ 25.59 seconds
Therefore, it takes approximately 25.59 seconds for the light-rail train to travel from one station to the next.