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
Two charged objects, A and B, are exerting an electric force on each other. What will happen if the charge on A is increased?
ozzi

Answer: The forces acting on both of them will increase in magnitude.

Explanation:

According to Coulomb's law, the electrostatic force between two bodies is proportional to the product of their two charges. If the charge on A is increased this product increases in size (it must have been non-zero to begin with, since there was a force between them at first). Thus, the force between them rises.

6 0
3 years ago
Calcular la aceleración que produce una fuerza de 40 N sobre un cuerpo con 88 Kg de masa. Expresar el resultado en m⁄s^2 *
tigry1 [53]

Answer:

a = 0.45 m/s²

Explanation:

The given question is ''Calculate the acceleration that produces a force of 40 N on a body with 88 kg of mass".

Given that,

Force, F = 40 N

Mass of the body, m = 88 kg

The net force acting on the body is given by :

F = ma

Where

a is the acceleration of the body

a=\dfrac{F}{m}\\\\a=\dfrac{40\ N}{88\ kg}\\\\a=0.45\ m/s^2

So, the required acceleration is 0.45 m/s².

4 0
3 years ago
The gap between electrodes in a spark plug is 0.060 cm. Producing an electric spark in a gasoline-air mixture requires an electr
Mandarinka [93]

Answer:

The magnitude of minimum potential difference is 1800 V

Explanation:

Given:

Electric field E = 3 \times 10^{6} \frac{V}{m}

Gap between electrodes d = 0.060 \times 10^{-2} m

For finding the minimum potential difference,

  \Delta V = E \times d

  \Delta V = 3 \times 10^{6} \times 0.060 \times 10^{-2}

  \Delta V = 1800 V

Therefore, the magnitude of minimum potential difference is 1800 V

8 0
3 years ago
A projectile enters a resisting medium at x = 0 with an initial velocity v0 = 910 ft/s and travels 5 in. before coming to rest.
evablogger [386]

Answer:

a = - 1.987 × 10⁶ ft/s²

t = 6.84 × 10⁻⁴ s

Explanation:

v₀ = 910 ft/s

x = 5 in.

relation v = v₀ - k x

v = 0 as body comes to rest

0 = 900 - 5k/12

k = 2184 s⁻¹

acceleration

\frac{\mathrm{d} v}{\mathrm{d} t} = -k\frac{\mathrm{d} x}{\mathrm{d} t}

where

(A) a = -k × v

 at v= 910 ft/s

     a = - 1.987 × 10⁶ ft/s²

(B)  at x = 3.9 in.

v = 910 - 3.9(2184)/12

v = 200.2 m/s

\frac{\mathrm{d} v}{\mathrm{d} t} = -k\frac{\mathrm{d} x}{\mathrm{d} t}

\frac{dv}{v} = -kdt

\int\limits^{200.2}_{900} {\frac{1}{v} }dv = -k\int\limits^t_0 dt

ln(200.2)-ln(900) = -kt

t = 6.84 × 10⁻⁴ s

3 0
2 years ago
An amusement park ride consists of a large vertical cylinder that spins about its axis fast enough that a person inside is stuck
ElenaW [278]

Let N be the normal force that forces the person against the wall.

Then u N = m g  is the frictional force supporting the person's weight

and N = m g / u

also, N = m v^2 / R is the normal force providing the centripetal acceleration

So, m g / u = m v^2 / R

v^2 = g R / u

since v = 2 pi R T

4 pi^2 R^2 T^2 = g R / u     and T^2 = g / (4  u pi^2 R)

T = 1/ (2 pi)  (g /(u R))^1/2 = .159 * (9.8 m/s^2 / (.521 * 4.4 m)) ^1/2

T = .68 / s

Do you see any thing wrong here?

T should have units of seconds not 1 / seconds

v should be  2 * pi * R / T  where T is the time for 1 revolution

So you need to make that correction in the above formula for v.

7 0
3 years ago
Other questions:
  • If a freely suspended vertical spring is pulled in downward direction and then released, which type of wave is produced in the s
    9·2 answers
  • A person doing chin-up weighs 700.0 N, disregarding the weight of the arms. During the first 25.0 cm of the lift, each arm exert
    8·2 answers
  • Differences Between light year and astronomical unit in two points .
    13·2 answers
  • 4. A latch holding a cart in place, hanging high above the ground, breaks and the cart falls. The cart started from rest and is
    7·1 answer
  • A person uses a constant force to push a 14.0 kg crate 1.80 m up a frictionless 10⁰ incline and to also increase its speed from
    14·1 answer
  • Rex Things throws his mother's crystal vase upwards with an initial velocity of 26 m/s. Determine the time it would take for the
    10·1 answer
  • Choose whether each of the following statements is true or false
    10·1 answer
  • While traveling on a dirt road, the bottom of a car hits a sharp rock anda small hole develops at the bottom of its gas tank. If
    11·1 answer
  • Fluid of intelligent​
    6·1 answer
  • Quick Quiz 40.2 While standing outdoors one evening, you are exposed to the following four types of electromagnetic radiation: y
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!