Train traveling between two stations. The train starts from rest and accelerates uniformly for 150 seconds. It then travels at a constant speed for 300 seconds and finally decelerates uniformly for 200 seconds.
Given that the distance between the two stations is 10450m, calculate the:
(a) Maximum speed, in km/h, the train attained;
(b) Acceleration, (c) Distance the train traveled during the last 100 seconds;
(d) Time the train takes to travel the first half of the journey.
If it accelerates at a m/s^2, then it reaches a top speed of 150a and travels 1/2 a*150^2 meters
Then it decelerates at 3/4 a for 200 seconds, and travels 150a*200 - 1/2 * 3/4 a * 200^2 meters
So, the total distance traveled is
1/2 a * 150^2 + 150a*300 + 150a*200 - 1/2 * 3/4 a * 200^2 = 10450
a = 11/75 m/s^2
See what you can do with that to answer the other bits.
To solve this problem, we need to break it down into three parts: the acceleration phase, the constant speed phase, and the deceleration phase.
(a) Maximum speed in km/h:
We know that the train accelerates uniformly for 150 seconds, travels at a constant speed for 300 seconds, and then decelerates uniformly for 200 seconds.
Let's first calculate the maximum speed attained during the acceleration phase. To do this, we can use the formula:
v = u + at
where v is the final velocity, u is the initial velocity (which is 0 in this case), a is the acceleration, and t is the time.
During the acceleration phase, the final velocity will be the maximum speed attained by the train.
Given: u = 0 m/s, t = 150 s
We need to find a.
Using the formula:
v = u + at
v = 0 + a(150)
v = 150a
Now, let's calculate the distance covered during the acceleration phase.
Using the formula:
s = ut + (1/2)at^2
Since the initial velocity (u) is zero, the formula simplifies to:
s = (1/2)at^2
Given: t = 150 s, s = 10450 m
Substituting the values, we get:
10450 = (1/2)a(150)^2
10450 = 11250a
a = 10450 / 11250
a = 0.928 m/s^2
Now, let's find the maximum velocity (v) using this acceleration (a).
v = 150a
v = 150 * 0.928 m/s
v ≈ 139.2 m/s
To convert the speed from m/s to km/h, we need to multiply by 3.6 (1 m/s = 3.6 km/h).
So, the maximum speed attained by the train is:
Max speed ≈ 139.2 * 3.6 km/h
Max speed ≈ 501.12 km/h
Therefore, the train attained a maximum speed of approximately 501.12 km/h.
(b) Acceleration:
We have already calculated the acceleration, which is approximately 0.928 m/s^2.
(c) Distance traveled during the last 100 seconds:
During the deceleration phase, the train decelerates uniformly for 200 seconds. We need to find the distance traveled during the last 100 seconds of this phase.
Using the formula:
s = ut + (1/2)at^2
Given: t = 100 s, a = -0.928 m/s^2 (negative because it's decelerating)
Substituting the values, we get:
s = 0 * 100 + (1/2) * (-0.928) * (100)^2
s = (-0.928) * 5000
s ≈ -4640 m
The distance traveled during the last 100 seconds is approximately 4640 m.
(d) Time taken to travel the first half of the journey:
The total time taken for the journey is 150 + 300 + 200 = 650 seconds.
To find the time taken to travel the first half of the journey, we divide the total time by 2.
Time taken to travel the first half = 650 / 2
Time taken to travel the first half = 325 seconds
Therefore, the train takes approximately 325 seconds to travel the first half of the journey.
To solve this problem, we will use the equations of motion for uniformly accelerated motion.
(a) Maximum speed:
To find the maximum speed, we need to calculate the final velocity when the train accelerates. Let's denote the initial velocity as u, the final velocity as v, the acceleration as a, and the time as t.
Using the formula for uniformly accelerated motion: v = u + at
Given: u = 0 (train starts from rest), t = 150 seconds, and a = acceleration
We need to find the final velocity, v.
Step 1: Calculate the acceleration.
The acceleration can be determined using the formula:
acceleration = change in velocity / time.
Given: time (t1) = 150 seconds, change in velocity (Δv) = v - u = v - 0 = v
acceleration = Δv / t1
acceleration = v / 150
Step 2: Calculate the final velocity.
Using the formula: v = u + at
v = 0 + (v / 150) * 150
v = v
This means that the final velocity when the train accelerates is the same as the maximum speed of the train.
(b) Acceleration:
We have already calculated the acceleration in step 1.
Acceleration = v / 150
(c) Distance traveled during the last 100 seconds:
During the last 100 seconds, the train decelerates uniformly. To find the distance traveled in this time period, we need to calculate the final velocity and use the formula for distance.
Given: Time (t2) = 200 seconds, deceleration = -acceleration (opposite direction of acceleration).
Step 1: Calculate the final velocity during deceleration.
Using the formula: v = u + at
v = v1 (velocity at the end of acceleration)
Step 2: Calculate the distance traveled during deceleration.
Using the formula for distance: s = ut + (1/2)at^2
s = v1 * t2 + (1/2) * (-acceleration) * t2^2
(d) Time to travel the first half of the journey:
The total distance between the two stations is known to be 10450 m. To find the time taken to travel the first half, we need to find the distance traveled during the first half and calculate the time using the formula for distance.
Given: Total distance = 10450 m
Step 1: Calculate the distance traveled during the first half.
Distance traveled during the first half = total distance / 2 = 10450 / 2
Step 2: Calculate the time taken to travel the first half.
Using the formula for distance: s = ut + (1/2)at^2
s = distance traveled during the first half, u = 0, a = acceleration, and t = time taken to travel the first half.
Substituting the values into the formula:
10450 / 2 = 0 * t + (1/2) * acceleration * t^2
Simplifying the equation and solving for t will give you the time taken to travel the first half of the journey.
Please note that for the calculations, you need to substitute the values of acceleration (calculated in step 1) and deceleration (opposite in sign to the acceleration value for the deceleration period) to get the final answers.