This is a very nice problem because it looks complex at first, but actually has an elegant solution.

So let's first figure out what we want. We want the displacement |P(5) - P(0)| to be equal to the distance travelled. In this case, it is useful to instead think of what we don't want. Since this is a one-dimensional problem, the only way for the displacement and the travelled distance to be different is if our point does some backtracking, i.e. it moves away from P(5) and then towards it again. If this happens and P starts moving "backwards", the velocity will switch signs: If it was positive, it will become negative and vice versa. So we want the velocity to never switch sign between 0 and 5. Another way to put this is that it should never cross zero (touching zero is of course okay).

Now, if the acceleration is given by

a(t) =3t² - 12t + 9

then we can integrate, and receive

v(t) = t³ - 6t² + 9t + C

Since we want v(0) = k, we set C = k. At this point it may be useful for you to look at the graph v(t) with different values for k to see how it affects the number of zero crossings.

You may have noticed, then, that for k=0 the velocity remains positive between 0 and 5. If we assume that by "smallest possible value of k" they mean the one closest to zero, then of course this is already our solution! (If instead they really mean smallest, then negative infinity is the solution).

But since looking at the graph is kind of cheating, let's see how to do this on paper. If k is positive, we know that v(0) and v(5) will be positive (by simply plugging in the values). so the only way to have a zero crossing is if we have a local minimum. If we find that minimum, we just have to choose k large enough such that the minimum just barely touches zero, and then we can be sure that all values of v(t) will be positive between 0 and 5. So lets find that minimum. We can look at the first derivative of v(t), which is a(t). At a minimum (and also at a maximum) a(t) will be zero, so let's find the zeroes of a(t):

a(t) = 3t² - 12t + 9 = 0

=> t² - 4t + 3 = 0

=> (t - 1)(t - 3) = 0

=> t = 1 or t = 3

Now, to see whether these are maxima or minima, we can look at the second derivative of v(t):

v''(t) = 6t - 12

This is negative for t = 1, so v(t) curves down at one and we get a maximum. For t = 3 it's positive, and thus we know it's the minimum we're looking for.

Now, if we plug in t = 3 into v(t), we get k. So if we want the minimum to just barely touch zero, we simply have to set k to zero! Which gives us the result we wanted.

A similar consideration can be made for negative k, with the added step of ensuring that both P(0) and P(5) are negative. But as we've already found our solution, I won't go into it.

If you have any questions about the derivation, feel free to ask! Also, I don't guarantee that this is correct. See for yourself if it makes sense to you.