Let's solve this kinematics problem step by step.

Given Data:

• Distance between stations X and Y: 2.05 km = 2050 m
• Train A accelerates from rest with an acceleration $a = 1 \, \text{m/s}^2$ for $t = 10 \, \text{s}$, and then continues at a uniform velocity.
• Train B travels at a constant velocity of $20 \, \text{m/s}$ towards station X.

a) Calculate the maximum speed that train A reaches after the first 10 s.

For Train A, we use the equation of motion for velocity under constant acceleration:

$v_f = v_i + at$ Where:

• $v_i = 0 \, \text{m/s}$ (initial velocity since the train starts from rest),
• $a = 1 \, \text{m/s}^2$ (acceleration),
• $t = 10 \, \text{s}$.

Thus,

$v_f = 0 + (1 \times 10) = 10 \, \text{m/s}.$

The maximum speed of Train A after 10 seconds is 10 m/s.

b) Determine the distance that train A covered during the first 10 s.

To calculate the distance traveled by Train A during the acceleration phase, we use the equation of motion:

$s = v_i t + \frac{1}{2} a t^2$

Substituting the values:

$s = 0 + \frac{1}{2} (1) (10^2) = \frac{1}{2} \times 100 = 50 \, \text{m}.$

So, Train A covers 50 meters during the first 10 seconds.

c) Determine the distance that train B covered during the first 10 s.

Since Train B is traveling at a constant velocity of 20 m/s, the distance covered by Train B in 10 seconds is:

$s = vt$

$s = 20 \times 10 = 200 \, \text{m}.$

So, Train B covers 200 meters during the first 10 seconds.

d) How long will it take the two trains to pass each other?

The initial distance between the two trains is 2.05 km = 2050 m. After 10 seconds:

• Train A has traveled 50 m,
• Train B has traveled 200 m.

The remaining distance between them after the first 10 seconds is:

$\text{Remaining distance} = 2050 - (50 + 200) = 1800 \, \text{m}.$

At this point, both trains are moving at constant velocities:

• Train A at 10 m/s,
• Train B at 20 m/s.

The relative speed between them is:

$\text{Relative speed} = 10 + 20 = 30 \, \text{m/s}.$

The time taken for the two trains to pass each other is:

$t = \frac{\text{Remaining distance}}{\text{Relative speed}} = \frac{1800}{30} = 60 \, \text{seconds}.$

So, it will take 60 seconds for the two trains to pass each other after the first 10 seconds.

Summary:

a) Maximum speed of Train A after 10 seconds: 10 m/s.
b) Distance covered by Train A in the first 10 seconds: 50 meters.
c) Distance covered by Train B in the first 10 seconds: 200 meters.
d) Time for the trains to pass each other after 10 seconds: 60 seconds.

Would you like further clarifications or detailed explanations on any step?

Related Questions:

1. What is the total distance covered by Train A by the time it passes Train B?
2. What would happen if Train A had an acceleration of 2 m/s² instead of 1 m/s²?
3. What would the time taken to pass be if Train B slowed down to 15 m/s after the first 10 seconds?
4. How would the time change if Train A decelerated after reaching its maximum velocity?
5. What would be the effect on the time if the distance between stations X and Y increased by 1 km?

Tip: Always ensure that the units are consistent when solving kinematics problems—especially between meters and kilometers, or seconds and hours!