Yes, good so far. I assume that you based this on the fact that the elevator is still accelerating up until 15 seconds.

One last thing that you can do, but which takes some time is finding the exact displacement x.

From W = mg, we know that the mass is 40 kg.

Hence, using F = ma, we know that the resultant acceleration is 12.5 m/s^2 by substituting F = 500 N and m = 40 kg. So, the acceleration of the elevator alone is = 12.5 - 10 = 2.5 m/s^2.

The distance covered can be obtained by s = ut + 0.5 at^2 and for 5 seconds, you get the displacement as 32.25 m.

Next part, we get resultant acceleration as 600/40 = 15 m/s^2, which means the elevator goes up at 5 m/s^2.

Now, we need to find the velocity after 10 seconds to determine the distance covered between 10 and 15 seconds.

v = u + at

We get the final velocity between 5 and 10 seconds as 12.5 m/s

Hence, the displacement for the second part is:

s = (12.5)(5) + 0.5(5)(5)^2 = 125 m

This is the additional distance covered.

I think you can finish this problem now

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If you don't like that, you can still make first an acceleration time graph by using F = ma.

When done, convert it into a velocity time graph by taking the area under the acceleration time graph. You get a straight line for each piece of 5 seconds.

Then, the area under the velocity time graph will give you the displacement time graph at any instant. I think it's easier, but more difficult to explain through a post.

I hope it helps!