Two trains face each other on adjacent tracks. They are initially at rest, and their front ends are 50 m apart. The train on the left accelerates rightward at 1.05 m/s2. The train on the right accelerates leftward at 0.93 m/s2.
(a) How far does the train on the left travel before the front ends of the trains pass?
(b) If the trains are each 150 m in length, how long after the start are they completely past one another, assuming their accelerations are constant?
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To solve this problem, we can break it down into two parts:
Part 1: Finding the distance traveled by the train on the left before the front ends of the trains pass.
Let's calculate the time taken by the trains to meet at the point where their distance is zero. We can use the kinematic equation:
s = ut + (1/2)at^2
where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time.
For the train on the left, initial velocity u is 0 m/s, acceleration a is 1.05 m/s^2, and we need to find the time taken t.
The distance traveled by the train on the left can be calculated using:
s = ut + (1/2)at^2
Rearranging the equation, we get:
t = sqrt(2s/a)
Since the distance s is given as 50 m and acceleration a is given as 1.05 m/s^2, we can substitute these values into the equation:
t = sqrt(2 * 50 m / 1.05 m/s^2)
t = sqrt(100 / 1.05)
t = sqrt(95.24)
t ≈ 9.76 s
So, the train on the left will travel for approximately 9.76 seconds before the front ends of the trains pass each other.
Part 2: Finding the time taken for the trains to be completely past each other.
To calculate this, we need to consider the lengths of the trains. The train on the left will need to travel the length of both trains (150 m + 150 m = 300 m) in addition to the initial 50 m. So, we need to find the time taken by the train on the left to travel a distance of 350 m (50 m + 300 m).
Using the same kinematic equation:
s = ut + (1/2)at^2
For the train on the left, initial velocity u is 0 m/s, acceleration a is 1.05 m/s^2, and we need to find the time taken t.
Using the equation, we have:
350 m = 0 * t + (1/2) * 1.05 m/s^2 * t^2
Rearranging the equation, we get:
0.525t^2 = 350 m
t^2 = 350 m / 0.525
t^2 ≈ 666.67
Taking the square root of both sides of the equation, we have:
t ≈ sqrt(666.67)
t ≈ 25.81 s
Therefore, it will take approximately 25.81 seconds for the trains to be completely past each other if their accelerations remain constant.
To summarize:
(a) The train on the left travels approximately 9.76 meters before the front ends of the trains pass.
(b) It takes approximately 25.81 seconds for the trains to be completely past each other, assuming constant acceleration.