A train travels between stations 1 and 2. The engineer of the train is instructed to start from rest at station 1 and accelerate uniformly between points A and B, then coast with a uniform velocity between points B and C, and finally accelerate uniformly between points C and D until the train stops at station 2. The distances AB, BC, and CD are all equal, and it takes 5.00 min to travel between the two stations. Assume that the uniform accelerations have the same magnitude, even when they are opposite in direction. How much of this 5.00 min period does the train spend b/w points A & B? B & C? C & D?

Ah, the train's journey is divided into three parts: A to B, B to C, and C to D. Let's give these parts some time in the spotlight!

Since the train starts from rest at station 1 and accelerates uniformly between points A and B, we can say that the train spends some time showcasing its acceleration skills in this segment. Let's call this time "t1".

Next up is the coasting part, where the train maintains a uniform velocity between points B and C. Let's call this time "t2".

Finally, the train accelerates uniformly between points C and D until it stops at station 2. We can give this time the name "t3".

Now let's divide these time segments fairly among the train's 5.00 min journey.

Since the train starts from rest, it needs half of the time to accelerate from point A to point B. Therefore, t1 = 5.00 min / 2 = 2.50 min.

For the coasting part, the train maintains its uniform velocity, so t2 takes the remaining time between A and D. With t2 = 5.00 min - 2.50 min = 2.50 min.

Lastly, the train accelerates from point C to point D, needing another half of the time remaining. That means t3 = 2.50 min / 2 = 1.25 min.

So, to summarize:

- The train spends 2.50 min between points A and B.
- The train spends 2.50 min between points B and C.
- The train spends 1.25 min between points C and D.

Now the train can carry on with its journey, armed with an understanding of its time distribution!

Let's assume the distance between points A and B is x.

Since the train starts from rest at station 1 and accelerates uniformly between points A and B, we can use the formula for displacement during constant acceleration:

s = ut + (1/2)at^2

where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.

At point A, the initial velocity u is 0, and the acceleration a is constant. The displacement between A and B is x, and we want to find the time it takes to travel this distance.

Using the formula, we have:

x = 0*t + (1/2)a*t^2
x = (1/2)a*t^2

Similarly, we can use the same formula for the displacement between points C and D:

x = 0*t + (1/2)a*t^2
x = (1/2)a*t^2

Now, to find the time spent between points A and B, we need to use the equation for the total time it takes to travel between the two stations:

5.00 min = time from A to B + time from B to C + time from C to D

Since the distances AB, BC, and CD are all equal, we can simplify the equation:

5.00 min = time from A to B + time from B to C + time from C to D
5.00 min = t + t + t
5.00 min = 3t

Thus, t = 5.00 min / 3 = 1.67 min

The train spends 1.67 min between points A and B.

Similarly, the train spends 1.67 min between points B and C, and also between points C and D.

So the breakdown of time spent is as follows:

Between points A and B: 1.67 min
Between points B and C: 1.67 min
Between points C and D: 1.67 min

To find out how much time the train spends between points A and B, between points B and C, and between points C and D, we can break down the journey into three parts.

Let:
- AB, BC, and CD represent the distances between the points A and B, B and C, and C and D, respectively.
- a represent the magnitude of the uniform acceleration.
- t1 represent the time taken to accelerate between points A and B.
- t2 represent the time taken to coast between points B and C.
- t3 represent the time taken to decelerate between points C and D.

Given that the distances AB, BC, and CD are all equal, we can assume they have the same value, let's say x.

To find t1, we can use the kinematic equation:
x = (1/2) * a * t1^2
Rearranging the equation, we get:
t1 = sqrt((2 * x) / a)

To find t2, we know that the velocity remains constant during coasting. Therefore, the distance traveled during this time is equal to the velocity multiplied by the time:
x = v * t2
Since the velocity remains constant, we can express it in terms of the initial velocity (0 at point A) and acceleration:
v = a * t1
Substituting the value of v, we get:
x = (a * t1) * t2
Simplifying, we find:
t2 = x / (a * t1)

To find t3, we can use the same equation used to find t1, since the same distance x is covered:
t3 = sqrt((2 * x) / a)

Now, let's calculate the values of t1, t2, and t3.

Given that it takes 5.00 minutes to travel between the two stations, the sum of t1, t2, and t3 should be equal to 5.00 minutes.

1. Calculate t1:
t1 = sqrt((2 * x) / a)

2. Calculate t2:
t2 = x / (a * t1)

3. Calculate t3:
t3 = sqrt((2 * x) / a)

To determine the time distribution between A & B, B & C, and C & D, substitute the values into the equation:
t1 + t2 + t3 = 5.00

After solving these equations simultaneously, you will determine the values of t1, t2, and t3.