Let's go through the questions step-by-step.

Given Data

• The distance between stations $X$ and $Y$ is $2.05 \, \text{km} = 2050 \, \text{m}$.
• Train A:
  • Starts from rest at station $X$.
  • Accelerates uniformly at $1.2 \, \text{m/s}^2$ for $10 \, \text{s}$.
  • After $10 \, \text{s}$, it travels at a uniform velocity.
• Train B:
  • Travels at a constant velocity of $20 \, \text{m/s}$ from station $Y$ towards station $X$.

Let's tackle each question.

Solution

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

Since Train A starts from rest and accelerates uniformly, we can use the formula: $v = u + at$ where:

• $u = 0 \, \text{m/s}$ (initial velocity),
• $a = 1.2 \, \text{m/s}^2$,
• $t = 10 \, \text{s}$.

Substitute the values: $v = 0 + (1.2)(10) = 12 \, \text{m/s}$

So, the maximum speed Train A reaches after 10 seconds is 12 m/s.

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

We use the formula for distance under constant acceleration: $s = ut + \frac{1}{2}at^2$ where:

• $u = 0 \, \text{m/s}$,
• $a = 1.2 \, \text{m/s}^2$,
• $t = 10 \, \text{s}$.

Substitute the values: $s = (0)(10) + \frac{1}{2}(1.2)(10)^2 = 0 + \frac{1}{2} \cdot 1.2 \cdot 100 = 60 \, \text{m}$

So, Train A covers a distance of 60 m in the first 10 seconds.

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

Since Train B moves at a constant velocity, we use the formula: $s = vt$ where:

• $v = 20 \, \text{m/s}$,
• $t = 10 \, \text{s}$.

Substitute the values: $s = (20)(10) = 200 \, \text{m}$

So, Train B covers a distance of 200 m in the first 10 seconds.

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

After 10 seconds:

• Train A is moving at a constant speed of 12 m/s.
• Train B continues at 20 m/s.
• The initial distance between them was 2050 m, but Train A covered 60 m, and Train B covered 200 m in the first 10 seconds. Thus, the remaining distance between them is: $2050 - 60 - 200 = 1790 \, \text{m}$

Now, the relative speed of Train A and Train B (since they are moving towards each other) is: $v_{\text{relative}} = 12 + 20 = 32 \, \text{m/s}$

The time it will take for them to meet is given by: $t = \frac{\text{distance}}{\text{relative speed}} = \frac{1790}{32} \approx 55.94 \, \text{s}$

So, it will take approximately 55.94 seconds for the two trains to pass each other.

Do you want further details or have any questions on these solutions?

Related Questions

1. How would the calculations change if Train A had a different acceleration rate?
2. What would happen if Train B started with a different initial speed?
3. How would we calculate the time to meet if Train A’s acceleration lasted longer than 10 seconds?
4. What is the impact of different initial distances between the stations on the meeting time?
5. How would the result change if Train B decelerated while approaching Train A?

Tip

Always ensure units are consistent when solving kinematics problems. Converting distances and velocities to the same units is essential for accuracy.