Exercise 3: Why Exact Velocity Doesn't Mean Euler's Method Is Perfectly Accurate — Possible Solution
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THE COLLEAGUE'S REASONING
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That because Euler's method computed velocity exactly right in this
chapter's own falling-object example, it must be a perfectly
accurate method overall.

WHY VELOCITY AND POSITION ARE GENUINELY DIFFERENT CASES HERE
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This chapter's own worked example specifically noted that velocity
matched EXACTLY, while position was off by 0.147 - these are two
separate results, not one combined "the method worked" outcome.
The reason velocity came out exact in that specific example is that
acceleration was CONSTANT (gravity, -9.8 m/s^2 the whole time) -
Euler's velocity update, v_new = v + a*dt, is exact whenever a is
constant, since it's just literally adding up a constant value dt
at a time, which is the same as ordinary multiplication (a*t) with
no approximation error at all.

Position is a genuinely different calculation - it depends on
VELOCITY, which is itself changing over time (accelerating). Using
the OLD velocity for each step's position update is itself an
approximation (this is exactly what Euler's method's own first-
order error comes from), which is why position error appeared even
though velocity had none.

WHY THIS DOESN'T GENERALIZE
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This chapter's own dt-vs-error verification showed the position
error shrinking linearly with dt (a genuine, real numerical error
that only vanishes as dt approaches zero) - this is a real
limitation of the method, not eliminated just because one specific
quantity (velocity, under constant acceleration specifically)
happened to be computed with zero error. If acceleration itself
were changing over time (not constant, as in most real physics
scenarios - air resistance, varying forces, etc.), even the
velocity calculation would develop its own first-order error, for
exactly the same reason position did here.

RESULT
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Euler's method is not "perfectly accurate" - it is a genuinely
useful but approximate numerical method with a real, verified,
first-order error. The exact velocity result in this chapter's own
example is a special case, specific to constant acceleration, not
evidence of the method's overall accuracy.

WHY THIS WORKS AS AN ANSWER
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The explanation identifies the SPECIFIC reason velocity happened to
be exact (constant acceleration making the velocity update a plain,
un-approximated sum) rather than treating it as a mysterious
property of the method itself, and explicitly connects back to this
chapter's own dt-vs-error verification to show the real limitation
still applies generally, including to velocity itself under
non-constant acceleration.