1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Akimi4 [234]
2 years ago
5

In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between st

ation stops. The more stops there are, the slower the train’s average speed. To get an idea of this problem, calculate the time it takes a train to make a 15.0-km trip in two situations: (a) the stations at which the trains must stop are 3.0 km apart (a total of 6 stations, including those at the ends); and (b) the stations are 5.0 km apart (4 stations total). Assume that at each station the train accelerates at a rate 1.1 m/s2 of until it reaches 95 km/h, then stays at this speed until its brakes are applied for arrival at the next station, at which time it decelerates at-2.0 m/s2 Assume it stops at each intermediate station for 22 s.
Physics
1 answer:
bekas [8.4K]2 years ago
7 0

Answer:

a) t = 746 s

b) t = 666 s

Explanation:

a)

  • Total time will be the sum of the partial times between stations plus the time stopped at the stations.
  • Due to the distance between stations is the same, and the time between stations must be the same (Because the train starts from rest in each station) we can find total time, finding the time for any of the distance between two stations, and then multiply it times the number of distances.
  • At any station, the train starts from rest, and then accelerates at 1.1m/s2 till it reaches to a speed of 95 km/h.
  • In order to simplify things, let's first to convert this speed from km/h to m/s, as follows:

       v_{1} = 95 km/h *\frac{1h}{3600s}*\frac{1000m}{1 km} = 26.4 m/s  (1)

  • Applying the definition of acceleration, we can find the time traveled by the train before reaching to this speed, as follows:

       t_{1} = \frac{v_{1} }{a_{1} } = \frac{26.4m/s}{1.1m/s2} = 24 s (2)

  • Next, we can find the distance traveled during this time, assuming that the acceleration is constant, using the following kinematic equation:

       x_{1} = \frac{1}{2} *a_{1} *t_{1} ^{2} = \frac{1}{2} * 1.1m/s2*(24s)^{2} = 316.8 m  (3)

  • In the same way, we can find the time needed to reach to a complete stop at the next station, applying the definition of acceleration, as follows:

       t_{3} = \frac{-v_{1} }{a_{2} } = \frac{-26.4m/s}{-2.2m/s2} = 12 s (4)

  • We can find the distance traveled while the train was decelerating as follows:

       x_{3} = (v_{1} * t_{3})   + \frac{1}{2} *a_{2} *t_{3} ^{2} \\ = (26.4m/s*12s) - \frac{1}{2} * 2.2m/s2*(12s)^{2} = 316.8 m - 158.4 m = 158.4m  (5)

  • Finally, we need to know the time traveled at constant speed.
  • So, we need to find first the distance traveled at the constant speed of 26.4m/s.
  • This distance is just the total distance between stations (3.0 km) minus the distance used for acceleration (x₁) and the distance for deceleration (x₃), as follows:
  • x₂ = L - (x₁+x₃) = 3000 m - (316.8 m + 158.4 m) = 2525 m (6)
  • The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       t_{2} = \frac{x_{2} }{v_{1} } = \frac{2525m}{26.4m/s} = 95.6 s   (7)

  • Total time between two stations is simply the sum of the three times we have just found:
  • t = t₁ +t₂+t₃ = 24 s + 95.6 s + 12 s = 131.6 s (8)
  • Due to we have six stations (including those at the ends) the total time traveled while the train was moving, is just t times 5, as follows:
  • tm = t*5 = 131.6 * 5 = 658.2 s (9)
  • Since we know that the train was stopped at each intermediate station for 22s, and we have 4 intermediate stops, we need to add to total time 22s * 4 = 88 s, as follows:
  • Ttotal = tm + 88 s = 658.2 s + 88 s = 746 s (10)

b)

  • Using all the same premises that for a) we know that the only  difference, in order to find the time between stations, will be due to the time traveled at constant speed, because the distance traveled at a constant speed will be different.
  • Since t₁ and t₃ will be the same, x₁ and x₃, will be the same too.
  • We can find the distance traveled at constant speed, rewriting (6) as follows:
  • x₂ = L - (x₁+x₃) = 5000 m - (316.8 m + 158.4 m) = 4525 m (11)
  • The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       t_{2} = \frac{x_{2} }{v_{1} } = \frac{4525m}{26.4m/s} = 171.4 s   (12)

  • Total time between two stations is simply the sum of the three times we have just found:
  • t = t₁ +t₂+t₃ = 24 s + 171.4 s + 12 s = 207.4 s (13)
  • Due to we have four stations (including those at the ends) the total time traveled while the train was moving, is just t times 3, as follows:
  • tm = t*3 = 207.4 * 3 = 622.2 s (14)
  • Since we know that the train was stopped at each intermediate station for 22s, and we have 2 intermediate stops, we need to add to total time 22s * 2 = 44 s, as follows:
  • Ttotal = tm + 44 s = 622.2 s + 44 s = 666 s (15)
You might be interested in
Imagine you had a bar of gold and decided to cut in half. You repeated this
MrMuchimi

Answer:

b, an element

Explanation:

this is because you never combined it with any other element in the periodic table and gold is not a compound or mixture.

7 0
3 years ago
Laura and Elana are discussing how to solve the following problem: "A canary sits 10 m from the end of a 30-m-long clothesline,
horrorfan [7]

Answer:

f=2.236\ Hz

Explanation:

Given:

Length of a rope,l=30\ m

Position of Canary on the rope from one end, l_c=10\ m

Position of Grackle on the rope from another end, l_g=5\ m

Tension in the rope, F_T=200\ N

linear mass distribution on the rope, \mu=0.1\ kg.m^{-1}

We have for the speed of wave on the string:

v^2=\frac{F_T}{\mu}

v^2=\frac{200}{\0.1}

v=44.7\ m.s^{-1}

<em>For canary to be undisturbed we need a node at this location.</em>

<em>Also, at the end close to Canary there must be a node to avoid any change in pattern of vibration.</em>

So,

the distance between Canary and the closer end must be equal to half the wavelength.

\frac{\lambda}{2} =10\ m

\Rightarrow \lambda=20\ m

∴Wavelength of wave to be produced = 20 m. This will give us nodes at the multiples of 10 and anti-nodes at the multiples of 5.

Now, frequency:

f=\frac{v}{\lambda}

f=\frac{44.7}{20}

f=2.236\ Hz

3 0
2 years ago
Which are two characteristics of a proton
Scilla [17]
Positive electric charge and found inside the nucleus
6 0
2 years ago
Read 2 more answers
The reaction is at dynamic equilibrium.
DiKsa [7]

Answer:

Nitrogen and hydrogen combine at the same rate that ammonia breaks down.

Explanation:

7 0
2 years ago
A softball is fouled off with a vertical velocity of 20 m/s and a horizontal velocity of 15 m/s. what is the resultant velocity
raketka [301]
25 m/s is the answer
8 0
3 years ago
Other questions:
  • PLEASE ANSWER QUICK!!! 2. Every magnet has _ unlike poles.
    7·1 answer
  • Disadvantage of intrapreneurship​
    14·1 answer
  • A cylindrical capacitor is made of two thin-walled concentric cylinders. The inner cylinder has radius 8 mm , and the outer one
    8·1 answer
  • Which method of testing substances is a sure way to identify a chemical reaction?
    8·2 answers
  • A) A 5.00-kg squid initially at rest ejects 0.250 kg of fluid with a velocity of 10.0 m/s. What is the recoil velocity of the sq
    8·1 answer
  • When white light is passed through a prism, it will
    6·2 answers
  • If a boy rides his bicycle 100 meters in 20 seconds to the end of the street how fast did he ride his bike
    10·1 answer
  • Which EM wave has the shortest wavelength?
    9·2 answers
  • Creative
    15·2 answers
  • What is the current in a circuit if the charge passing each point is 10current <br>in 2s​
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!