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?
I'm not sure how I'm supposed to set up the equation. I know avg v= delta x over delta t, but I can't get it so that it works.
The average V while accelerating and decelerating is half the uniform velocity between B and C. If all three distances are equal, it will therefore take twice as long to do AB or CD as it will to travel BC. That means 2/5 of the time is spent accelerating and decerlating, and 1/5 at constant velocity (BC).
You had the right equation in mind, but just needed to follow through with the logic.
Therefore, the train spends 2/5 of the 5.00 min period between points A and B, 1/5 between points B and C, and 2/5 between points C and D.
To solve this problem, let's assign variables to the distances AB, BC, and CD and the time spent in each section.
Let's say the distance AB, BC, and CD are all equal and we can call it 'd'.
Let's represent the time spent in the section AB as 't1', the time spent in the section BC as 't2', and the time spent in the section CD as 't3'.
Given that it takes 5.00 min to travel between the two stations, we can write the equation: t1 + t2 + t3 = 5.00 min.
Now, let's consider the uniform acceleration.
During acceleration from rest at A to B, the average velocity is half the final velocity. Let's represent the final velocity at B as 'v'.
During the time spent in AB (t1), the train accelerates uniformly. Using the equation v = u + at, where 'u' is the initial velocity, 'a' is the acceleration, and 't' is the time, we can rewrite it as vf (velocity at B) = v0 (initial velocity at A) + a(t1).
Similarly, during deceleration from C to D, the average velocity is also half the final velocity. Let's represent the final velocity at C as 'v'.
During the time spent in CD (t3), the train decelerates uniformly. Using the equation vf = v0 + at, we can rewrite it as vf (velocity at D) = v0 (initial velocity at C) - a(t3).
Now, let's consider the time spent in BC (t2).
During the time spent in BC (t2), the velocity remains constant. Let's call it 'v2'.
Now, we know that the average velocity while accelerating and decelerating is half the constant velocity between B and C.
Therefore, we can write: (v + v2) / 2 = v2 / 2.
Now, let's substitute the variables into the equation t1 + t2 + t3 = 5.00 min and solve for t1, t2, and t3.
From the given information, we know that t2 (time spent in BC) is equal to 1/5th of the total time. Therefore, t2 = (1/5) * 5.00 = 1.00 min.
Since the distances AB, BC, and CD are all equal, and it takes twice as long to travel AB or CD as it does to travel BC, we can conclude that t1 and t3 are equal. Therefore, t1 = t3.
Substituting the variable values into the equation t1 + t2 + t3 = 5.00 min, we get:
t1 + 1.00 + t1 = 5.00
2t1 = 4.00
t1 = 2.00 min
Therefore, the train spends 2.00 min between points A and B, 1.00 min between points B and C, and 2.00 min between points C and D.
To set up the equation, let's first denote the time it takes to accelerate from A to B as t1, the time it takes to coast from B to C as t2, and the time it takes to decelerate from C to D as t3. Since the distances AB, BC, and CD are all equal, let's denote the distance between each point as x.
Now, let's consider the time it takes for the train to travel from A to B. The train starts from rest at A and accelerates uniformly. The average velocity during this time is half the final velocity (since acceleration is uniform), and the average velocity is given by the distance travelled divided by the time taken. So we have:
Average velocity from A to B = x / t1
Since we know the total time taken from A to D is 5 minutes, we can set up the equation:
t1 + t2 + t3 = 5
Now, let's consider the time it takes for the train to travel from B to C. The train is coasting at a uniform velocity, so the average velocity during this time is just the velocity at B. Since the distance BC is equal to x, the average velocity from B to C is x / t2.
Finally, let's consider the time it takes for the train to travel from C to D. The train decelerates uniformly, so the average velocity during this time is again half the final velocity. Using the same logic as before, we have:
Average velocity from C to D = x / t3
Now we have three equations:
Average velocity from A to B = x / t1
Average velocity from B to C = x / t2
Average velocity from C to D = x / t3
And we also have the equation:
t1 + t2 + t3 = 5
From here, you can solve the system of equations to find the values of t1, t2, and t3, which will give you the amount of time the train spends between points A and B, B and C, and C and D, respectively.