Two trains face each other on adjacent tracks. They are initially at rest, and their front ends are 30 m apart. The train on the left accelerates rightward at 1.17 m/s^2. The train on the right accelerates leftward at 1.1 m/s^2.

(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?

Please use kinematics equations to solve.

D1 = Distance traveled by train on left.

D2 = Distance traveled by train on right

a. D1 + D2 = 30 m.
0.5a1*t^2 + 0.5a2*t^2 = 30
0.5*1.17*t^2 + 0.5*1.1*t^2 = 30
0.585t^2 + 0.55t^2 = 30.
1.135t^2 = 30.
t^2 = 26.43
t = 5.14s.

D1 = 0.5*a1*t^2.
a1 = 1.17 m/s^2.
t = 5.14 s.
Solve for D1.

b. D1 - D2 = 300 m.
0.5*1.17*t^2 + 0.5*1.1*t^2 = 2*150 = 300
0.585t^2 + 0.55t^2 = 300
1.135t^2 = 300
t^2 = 264.3
t = 16.3 s.

Correction:

b. D1 + D2 = 30+300 = 330 m.
0.5*1.17*t^2 + 0.5*1.1*t^2 = 330.
0.585t^2 + 0.55t^2 = 330.
1.135t^2 = 330.
t^2 = 290.75.
t = 17.1 s.

Alright, let's put on our mathematical hats and solve this problem using kinematics equations. But don't worry, I'll still throw in some humor along the way!

(a) To find the distance traveled by the train on the left before the front ends pass, we can use the equation:

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

Since the train is initially at rest (u = 0 m/s) and the acceleration is given as 1.17 m/s^2, we can simplify the equation to:

s = (1/2)at^2

Plugging in the values, we have:

s = (1/2)(1.17 m/s^2)t^2

To find the time it takes for the trains to pass, we can use the equation:

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

Since the trains are facing each other, their relative velocity is the sum of their individual accelerations:

v = 1.17 m/s^2 + (-1.1 m/s^2) = 0.07 m/s^2

Now, let's substitute this relative velocity into the equation:

30 m = 0.07 m/s^2 * t

Solving for t, we get:

t = 30 m / 0.07 m/s^2 = 428.57 s

Now we can substitute this value of t back into the equation for distance traveled by the train on the left:

s = (1/2)(1.17 m/s^2)(428.57 s)^2

Calculating this out, we get:

s ≈ 27704.76 m

So, the train on the left travels approximately 27704.76 meters before the front ends pass.

(b) To find the time it takes for the trains to be completely past each other, we can use the equation:

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

Since the trains are moving in opposite directions, their relative velocity is the sum of their individual accelerations:

v = 1.17 m/s^2 + 1.1 m/s^2 = 2.27 m/s^2

Now, let's substitute this relative velocity into the equation:

300 m = 2.27 m/s^2 * t

Solving for t, we get:

t = 300 m / 2.27 m/s^2 ≈ 132.16 s

So, it takes approximately 132.16 seconds for the trains to be completely past each other.

Voila! I hope you enjoyed this kinematic journey with me. Remember, laughter is always the best acceleration in life!

To solve this problem, we can use the kinematic equations of motion. Let's go step by step to find the answers to both parts of the question:

(a) How far does the train on the left travel before the front ends of the trains pass?

We can use the equation for displacement in terms of time, acceleration, and initial velocity:
s = ut + (1/2)at^2

In this case, the train on the left starts from rest, so its initial velocity (u) is 0. The acceleration (a) is given as 1.17 m/s^2.

Since the trains face each other, their relative velocity is the sum of their individual velocities. The train on the left is moving rightward, so the relative velocity is the velocity of the left train subtracted from the velocity of the right train, which is 0 - (-1.1) = 1.1 m/s.

We want to find the distance (s) travelled by the left train before the front ends of the trains pass. The time (t) can be calculated using the equation:
v = u + at

For the left train:
1.1 = 0 + 1.17t

Solving for t:
t = 1.1 / 1.17
t ≈ 0.940 seconds

Now we can calculate the distance travelled by the left train using the equation for displacement:
s = ut + (1/2)at^2
s = 0 + (1/2)(1.17)(0.940)^2
s ≈ 0.512 meters

Therefore, the train on the left travels approximately 0.512 meters 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?

To find the time required for the trains to completely pass each other, we need to calculate the time it takes for the rear end of the right train to reach the front end of the left train.

Let's start by finding the time (t1) to travel the initial distance of 30 meters between the front ends of the trains. We can use the equation:
s = ut + (1/2)at^2

For the right train (which is moving leftward):
30 = 0 + (1/2)(-1.1)t1^2

Solving for t1:
t1^2 ≈ (2*30) / (-1.1)
t1 ≈ √(60 / (-1.1))

Next, we need to calculate the time (t2) for the distance covered by the rear end of the right train to pass the front end of the left train. This distance is the sum of the length of the left train (150 m) and the initial distance between the front ends (30 m).

For the right train:
180 = 0 + (1/2)(-1.1)t2^2

Solving for t2:
t2^2 ≈ (2*180) / (-1.1)
t2 ≈ √(360 / (-1.1))

Now, we can find the total time (t_total) required for the trains to completely pass each other by adding t1 and t2:
t_total = t1 + t2

Therefore, the time required for the trains to completely pass each other, assuming their accelerations are constant, is approximately t_total.