A subway train is designed with a maximum acceleration of ±0.637 m/s2 , which allows for both passenger safety and comfort.
(a) If subway stations are 8.6 km apart, what is the maximum velocity that can be obtained between stations?
(b) How long does it take to travel between two stations?
(c) The train stops for a total of 50 s at each station. What is the overall average velocity of the train from station to station?
I got part a right with an answer of 74.01 m/s.
When I did parts b & c, my answers were incorrect.
Any help would be greatly apreciated!
Distance travelled while accelerating = 4.3 km
Vmax = sqrt(2aX) = 74.01 m/s
Average velocity = 37.0 m/s
Total travel time between stops
= 8600 m/37.0 m/s = 232 s
Average velocity including one stop:
(8600 m)/(232 + 50) = 30.5 m/s
To solve parts (b) and (c), let's break down the problem using a few physics equations.
(a) Maximum velocity can be obtained using the formula:
V_max = sqrt(2 * a_max * d)
where V_max is the maximum velocity, a_max is the maximum acceleration (+0.637 m/s²), and d is the distance between stations (8.6 km = 8600 m).
Plugging in the values:
V_max = sqrt(2 * 0.637 m/s² * 8600 m)
V_max ≈ 74.01 m/s
Your answer for part (a) is correct.
Now let's move on to part (b) and calculate the time it takes to travel between two stations.
(b) To find the time, we need to know the time it takes to accelerate the train from rest to its maximum velocity (t1), as well as the time it takes to decelerate from maximum velocity to rest (t2).
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.
For acceleration (t1):
0 m/s (initial velocity) to V_max (final velocity),
u = 0 m/s
V = V_max ≈ 74.01 m/s
a = a_max = +0.637 m/s²
V = u + at
74.01 m/s = 0 + (0.637 m/s²) * t1
Solving for t1:
t1 ≈ 116.06 s
For deceleration (t2):
V_max (initial velocity) to 0 m/s (final velocity),
u = V_max ≈ 74.01 m/s
V = 0 m/s
a = -a_max = -0.637 m/s² (negative because it's decelerating)
V = u + at
0 = V_max + (-0.637 m/s²) * t2
Solving for t2:
t2 ≈ 116.06 s
The total time to travel between two stations (t_total) is the sum of t1, t2, and the time spent at each station (50 s):
t_total = t1 + t2 + 50 s
t_total ≈ 116.06 s + 116.06 s + 50 s
t_total ≈ 282.12 s
So, the time it takes to travel between two stations is approximately 282.12 seconds.
Next, let's move on to part (c) and calculate the overall average velocity of the train from station to station.
(c) The overall average velocity can be calculated using the formula:
Average velocity = Total distance / Total time
where Total distance is the distance between two stations (8.6 km = 8600 m), and Total time is the time calculated in part (b) (282.12 s).
Plugging in the values:
Average velocity = 8600 m / 282.12 s
Average velocity ≈ 30.53 m/s
So, the overall average velocity of the train from station to station is approximately 30.53 m/s.
I hope this helps! Let me know if you have any further questions.