We are given the following information:

• Initial velocity, $u = 10 \, \text{m/s}$
• Constant acceleration, $a = 3 \, \text{m/s}^2$
• Distance traveled, $s = 60 \, \text{m}$

We need to find the final velocity $v$ of the train after it has traveled 60 m.

We can use the third equation of motion to solve for $v$:

$v^2 = u^2 + 2as$

Substitute the known values:

$v^2 = (10)^2 + 2 \times 3 \times 60$ $v^2 = 100 + 360$ $v^2 = 460$

Now, take the square root of both sides to find $v$:

$v = \sqrt{460} \approx 21.47 \, \text{m/s}$

Thus, the final velocity of the train when it crosses 60 m is approximately 21.47 m/s.

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Would you like to see the detailed steps, or do you have any other questions? Here are some related questions to explore further:

1. How would the result change if the acceleration were different?
2. What is the time it would take for the train to travel 60 meters?
3. How can you derive the equations of motion from basic principles?
4. How does the initial velocity impact the final velocity in this problem?
5. What would be the total distance covered if the train were to continue with the same acceleration for a longer period?

Tip: The third equation of motion $v^2 = u^2 + 2as$ is especially useful when you need to find velocity after a certain distance without having time.