A 15,000-kg train accelerates uniformly from 0 to 15 m/s when passing through a distance of 150 m. What is its kinetic energy when it is 75 m down the track?

Question

theeric · Answer

Have you attempted to arrive at your own solution yet? This is an interesting and slightly complex problem!

theeric · Answer

If it seems overwhelming at first, I suggest that you calm down and look at the most basic thing until it's fresh in your head.

For starters, what does it mean that the train accelerates? Acceleration is a change in velocity. So the train's velocity is changing! But wait, the train $\sf accelerates\ uniformly$. What does that mean?

Now you can pick up from there!

If you start working through and find that you have a more specific question, then feel free to ask! If you have trouble finding the way to get a solution, you can just continue examining the problem. If that doesn't work, then feel free to ask for a hint or a way to analyze a question like this.

I might not be on later, but you can always bump the question!

Relevant equations
(they're physics, spoken in the math language):

$\color{green}{KE}=\dfrac{1}{2}mv^2$
$v_f^2=v_i^2+2ad$

I put kinetic energy in $\sf\color{green}{green}$ because the question asks:
"What is its $\sf\color{green}{kinetic\ energy}$ when it is 75 m down the track?"

So it's your goal. You want to look through everything and eventually be able to solve for $\color{green}{KE}$.