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?
The distance covered is 50 m, so let the rightmost train go x meters, the left goes 50-x m in the same time...
x=1/2 aright*time^2
50-x=1/2 aleft*time^2
divide the equations
x/(50-x)=aright/aleft
solve for x (right), and 50-x (left)
b. now, the distance apart the ends starts at 350m.
one goes y, the other 350-y
solve it again, only this part find the time.
To solve this problem, we need to analyze the motion of both trains and determine when their front ends pass each other.
Let's start by finding the time it takes for the front ends of the trains to pass each other.
For train A (the train on the left), we can use the kinematic equation:
d = v₀t + 1/2at^2
Since the train starts from rest, its initial velocity (v₀) is 0. The acceleration (a) is given as 1.05 m/s². The distance covered (d) is the initial separation between the trains, which is 50 m.
Plugging in the values, we have:
50 = 0 + 1/2 * 1.05 * t^2
Rearranging the equation, we get:
25 = 0.525t^2
Now, solve for t:
t^2 = 25 / 0.525
t^2 ≈ 47.62
t ≈ √47.62
t ≈ 6.91 seconds (rounded to two decimal places)
So, it takes approximately 6.91 seconds for the front ends of the trains to pass each other.
Now, let's move on to part (a), finding how far the train on the left travels before the front ends pass each other.
Using the kinematic equation again, but this time solving for d:
d = v₀t + 1/2at^2
Since the initial velocity (v₀) is still zero, we have:
d = 0 + 1/2 * 1.05 * (6.91)^2
d ≈ 0 + 1/2 * 1.05 * 47.62
d ≈ 24.96 meters (rounded to two decimal places)
Therefore, the train on the left travels approximately 24.96 meters before the front ends of the trains pass each other.
Moving on to part (b), determining when the trains are completely past each other.
Since each train is 150 meters long, we add the length of both trains to the initial separation between them:
Total length = 150 + 150 + 50 = 350 meters
Now, we need to find the time it takes for a train to travel this distance. We can use the equation:
d = v₀t + 1/2at^2
For train B (the train on the right), the initial velocity (v₀) is 0, and the acceleration (a) is -0.93 m/s² (negative because it's in the opposite direction).
350 = 0 + 1/2 * -0.93 * t^2
Simplifying the equation:
175 = -0.465t^2
t^2 = 175 / -0.465
t^2 ≈ -376.34
Since we can't have a negative time, it means the trains will never be completely past each other.
In conclusion, the train on the left travels approximately 24.96 meters before the front ends of the trains pass each other. However, they will never be completely past each other since their combined length is greater than the initial separation between them.
To solve this problem, we will use the equations of motion to calculate the distance traveled and the time required for the trains to pass each other.
(a) To find the distance the train on the left travels before the front ends of the trains pass, we need to find the time it takes for the front ends to meet.
First, let's find the time it takes for the train on the left to reach the same speed as the train on the right:
Using the equation `v = u + at`, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time elapsed:
For the train on the left:
Initial velocity, u = 0 m/s
Acceleration, a = 1.05 m/s^2
Substituting these values into the equation, we get:
v = 0 + (1.05 m/s^2)t
Since the final velocity of the train on the left when the front ends meet is the same as the initial velocity of the train on the right, we can set them equal to each other:
(1.05 m/s^2)t = 0.93 m/s^2 * t
Simplifying the equation:
1.05t = 0.93t
0.12t = 0
t = 0
This means that the train on the left never reaches the same speed as the train on the right, so the train on the left will never pass the front end of the train on the right. Therefore, the train on the left will travel an infinite distance before the front ends of the trains meet.
(b) To find the time it takes for the trains to completely pass each other, we will consider the motion of the train on the right.
The train on the right starts from rest, so its initial velocity is 0 m/s. The final velocity can be found using the equation `v = u + at`, where u is the initial velocity, a is the acceleration, t is the time elapsed, and v is the final velocity:
v = 0 + (0.93 m/s^2)t
The final velocity of the train on the right when the front ends meet is the sum of its initial velocity and the velocity it acquires during the time it takes to meet the train on the left. So we have:
v = 0 + (0.93 m/s^2)t
Now, let's find the time it takes for the front ends to meet at the point where they pass each other. The distance traveled by the front end of the train on the right is the sum of the initial separation of 50 m and the length of the train on the right, which is 150 m. So, the distance traveled is 200 m.
Using the 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 elapsed, we can solve for t:
s = ut + (1/2)at^2
200 m = 0 + (1/2)(0.93 m/s^2)t^2
Now, we solve for t:
200 m = (0.465 m/s^2)t^2
t^2 = (200 m) / (0.465 m/s^2)
t^2 = 430.108 m^2/s^2
t ≈ 20.74 s
So, it takes approximately 20.74 seconds for the trains to completely pass each other, assuming their accelerations are constant.