The speed is not linear; it is not constant; it is not average.

v = 10t -t^2

That is a parabola. A vertical parabola that opens downward because of the negative t^2.

That is because the particle keeps on accelerating up to the mid-distance between A and B, and then it keeps on decelerating right after that mid-distance. (The acceleration and deceleration are equal and opposite and constant.....that's why a parabola.)

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The vertex of that parabola is the maximum speed, and it happens at the mid-distance between A and B.

There are many ways to find the vertex. (Completing the square, or byCalculus(dv/dt = 0), or by formula "x = -b/2a"

You should find it at (10,25) .....or, at 10 seconds, and max. V = 25 m/sec.

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But for the purpose of this qn, the vertex is not needed.

What are the t's when v = 5 m/sec?

v = 10t -t^2

5 = 10t -t^2

t^2 -10t +5 = 0

t = {10 +,-sqrt[10^2 -4(1)(5)]} / 2(1)

t = {10 +,-4sqrt(5)} /2

t = 5 +,-2sqrt(5) ...........in seconds.

So, t1 = 5 -2sqrt(5) sec

t2 = 5 +2sqrt(5) sec.

That means, see the parabola, between t1 and t2, inclusive, the v is equal or higher than 5 m/sec.

How long is that in seconds?

Why, from t1 to t2, of course.

And that is 4sqrt(5) seconds.