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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.

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### 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.

Two railway stations are 2.05 km apart. Train A travels from rest from station X and accelerates uniformly at 1.2 m/s^2 for 10 s in the direction of station Y. After the 10 s, train A then travels further at a uniform velocity. At the same time that A departs, train B travels at a uniform velocity of 20 m/s in the direction of station X. Train B maintains this velocity.

Questions:

a) Calculate the maximum speed that train A reaches after the first 10 s.
b) Determine the distance that train A covered during the first 10 s.
c) Determine the distance that train B covered during the first 10 s.
d) How long will it take the two trains to pass each other?

Solve a kinematics problem involving two trains, where Train A accelerates from rest with 1.2 m/s^2 for 10 seconds and Train B travels at a constant speed of 20 m/s. Calculate maximum speed, distance covered by each train in the first 10 seconds, and the time for both trains to meet. Suitable for Grade 11 students.

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