https://file.aiquickdraw.com/mathbot/file/1728400136020sz5u9pkd.jpg

Let's break down the problem from the image.

### Information provided:
- **Distance between two stations (X and Y):** 2.05 km (2050 m)
- **Train A:** Starts from rest and accelerates uniformly at \( 1 \, \text{m/s}^2 \) for 10 seconds.
- After 10 seconds, Train A travels at a constant velocity.
- **Train B:** Travels at a uniform velocity of 20 m/s in the opposite direction (from station Y to station X).

Now, let's address the questions one by one:

### a) Calculate the maximum speed that train A reaches after the first 10 seconds.
Train A starts from rest, and accelerates at \( 1 \, \text{m/s}^2 \).

Using the equation of motion for velocity:
\[
v = u + at
\]
Where:
- \( u = 0 \, \text{m/s} \) (initial velocity),
- \( a = 1 \, \text{m/s}^2 \) (acceleration),
- \( t = 10 \, \text{seconds} \).

Thus:
\[
v = 0 + (1 \times 10) = 10 \, \text{m/s}
\]
**Train A's maximum speed after 10 seconds is 10 m/s.**

### b) Determine the distance that train A covered during the first 10 seconds.
The equation for distance covered under uniform acceleration is:
\[
s = ut + \frac{1}{2} a t^2
\]
Where:
- \( u = 0 \, \text{m/s} \),
- \( a = 1 \, \text{m/s}^2 \),
- \( t = 10 \, \text{seconds} \).

Thus:
\[
s = (0 \times 10) + \frac{1}{2} \times 1 \times 10^2 = 0 + \frac{1}{2} \times 100 = 50 \, \text{meters}
\]
**Train A covers 50 meters during the first 10 seconds.**

### c) Determine the distance that train B covered during the first 10 seconds.
Train B is moving at a constant speed of 20 m/s.

The distance covered by Train B in 10 seconds is:
\[
s = vt = 20 \times 10 = 200 \, \text{meters}
\]
**Train B covers 200 meters during the first 10 seconds.**

### d) How long will it take the two trains to pass each other?
The total distance between the two trains is 2050 meters. After 10 seconds:
- Train A has moved 50 meters.
- Train B has moved 200 meters.

Thus, after 10 seconds, the remaining distance between the two trains is:
\[
2050 - (50 + 200) = 2050 - 250 = 1800 \, \text{meters}
\]

Now, Train A is moving at a constant velocity of 10 m/s, and Train B is moving at 20 m/s. The relative speed between the two trains is:
\[
10 + 20 = 30 \, \text{m/s}
\]

The time taken for them to meet can be calculated using:
\[
t = \frac{\text{remaining distance}}{\text{relative speed}} = \frac{1800}{30} = 60 \, \text{seconds}
\]
**It will take the two trains 60 seconds to pass each other after the first 10 seconds.**

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Would you like more details or have any questions?

### Related Questions:
1. What is the total distance covered by Train A before it meets Train B?
2. How far has Train B traveled in total before passing Train A?
3. What is the acceleration formula, and how is it used in different motion problems?
4. How would the time change if Train A had a higher acceleration?
5. How would you calculate the distance between the two trains after 5 seconds?

### Tip:
For problems involving uniform acceleration, always remember to apply the correct kinematic equations based on the known quantities (initial velocity, acceleration, and time).

Two railway stations are 2.05 km apart. Train A travels from rest from station X and accelerates uniformly at 1 m/s² for 10 s in the direction of station Y. After the 10 s, train A then travels further at a uniform velocity. Train B travels at a uniform velocity of 20 m/s in the direction of station X. The tasks are to calculate: a) maximum speed that train A reaches after 10 s, b) distance train A covers during the first 10 s, c) distance train B covers during the first 10 s, and d) how long it takes the two trains to pass each other.

Solve this kinematics problem where two trains travel with different motions: one accelerates, while the other moves at constant velocity. Learn how to calculate speed, distance, and the time for the two trains to pass each other. Suitable for Grade 11 students.

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