A train normally travels at a uniform speed of 74km/h on a long stretch of straight, level track. On a particular day, the train must make a 3.0min stop at a station along this track.
If the train decelerates at a uniform rate of 1.3m/s2 and, after the stop, accelerates at a rate of 0.70m/s2 , how much time is lost because of stopping at the station?
Vo = 74km/h = 74000m/3600s = 20.56 m/s.
V = Vo + a*t
t1 = (V-Vo)/a = (0-20.56)/-1.3 = 15.8 s.
t2 = 3 min = 180 s.
t3 = 20.56-0)/0.7 = 29.37 s
T=t1 + t2 + t3=15.8 + 180 + 29.37=225.2s
= 3.75 min. = Time lost.
To calculate the time lost due to stopping at the station, we need to find the time it takes for the train to decelerate to a stop and then accelerate back to its original speed.
1. First, let's find the distance required to decelerate from the initial speed of 74 km/h to a complete stop. We'll need to convert the speed from km/h to m/s for consistent units.
Speed = 74 km/h = 74,000 m/60 min ÷ 60 sec = 20.56 m/s
2. We can use the formula for uniformly decelerating motion to find the stopping distance:
v^2 = u^2 - 2as
where v is the final velocity (0 m/s), u is the initial velocity (20.56 m/s), a is the acceleration (-1.3 m/s^2), and s is the stopping distance.
Solving for s:
0^2 = 20.56^2 - 2(-1.3)s
0 = 421.77 + 2.6s
-2.6s = 421.77
s = -421.77 ÷ -2.6 ≈ 162.22 m
Therefore, the stopping distance is approximately 162.22 meters.
3. Next, let's find the time it takes for the train to decelerate to a stop.
We can use the formula:
v = u + at
where v is the final velocity (0 m/s), u is the initial velocity (20.56 m/s), a is the acceleration (-1.3 m/s^2), and t is the time.
Solving for t:
0 = 20.56 + (-1.3)t
1.3t = 20.56
t = 20.56 ÷ 1.3 ≈ 15.81 s
Therefore, the time taken to decelerate to a stop is approximately 15.81 seconds.
4. After the stop, the train needs to accelerate back to its initial speed of 74 km/h. We can use a similar calculation to find the time taken to accelerate.
5. First, let's find the distance required to accelerate from a complete stop back to the initial speed.
Using the same formula as above, but with a positive acceleration now, we can find the distance:
v^2 = u^2 + 2as
Since the final velocity is still 74 km/h after decelerating to a stop, we'll now convert it to m/s:
Speed = 74 km/h = 74,000 m/60 min ÷ 60 sec = 20.56 m/s
Solving for s:
(20.56)^2 = 0^2 + 2(0.7)s
421.77 = 1.4s
s = 421.77 ÷ 1.4 ≈ 301.27 m
Therefore, the distance required to accelerate from a complete stop back to the initial speed is approximately 301.27 meters.
6. Next, let's find the time it takes for the train to accelerate back to the speed of 74 km/h (20.56 m/s).
Again, using the formula:
v = u + at
We can solve for t:
20.56 = 0 + 0.7t
0.7t = 20.56
t = 20.56 ÷ 0.7 ≈ 29.37 s
Therefore, the time taken to accelerate back to the initial speed is approximately 29.37 seconds.
7. Finally, we can calculate the total time lost due to stopping at the station.
The total time lost is the sum of the deceleration time and the acceleration time:
Total time lost = Deceleration time + Acceleration time
Total time lost = 15.81 s + 29.37 s ≈ 45.18 s
Therefore, the time lost because of stopping at the station is approximately 45.18 seconds.