How fast can you turn around?

There are two great solutions here already, so just adding the visual approach using velocity v/s time graph - the area under which is the distance covered.

Imgur

Consider three areas or distances covered

A1 - covered while speeding up

A2 - covered while slowing down

A3 - covered in the return journey

A1 = A2 . . . congruent right triangles <=identical height and slope of hypotenuse

For a round trip, A3 = A1 + A2 = 2 A2

Because they are both right angled triangles and share a pair of vertically opposite angles, the blue and the red triangles are similar.

Similar triangles with areas 1:2 must have sides $t : t' = 1 : \sqrt{2}$ Hence $T = t + t + t' = (2 + \sqrt{2}) t = \boxed { 3.414 t}$

After time $t$ the particle will have traveled a distance $(\frac{1}{2})at^{2}$ and be traveling at a speed of $at$ .

Now reset the clock to $t_{1} = 0$ . The acceleration is now $-a$ , and with an initial speed of $at$ the distance $D$ from the original starting point is given by the formula

$D = (\frac{1}{2})at^{2} + att_{1} - (\frac{1}{2})at_{1}^{2}$ .

Now the particle will have returned to its starting point when $D = 0$ , i.e., when

$(\frac{1}{2})at_{1}^{2} - att_{1} - (\frac{1}{2})at^{2} = 0 \Longrightarrow t_{1}^{2} - tt_{1} - t^{2} = 0$

$\Longrightarrow t_{1} = \dfrac{2t \pm \sqrt{4t^{2} + 4t^{2}}}{2} = (1 + \sqrt{2})t$ ,

where we took the positive root since we must have $t_{1} \gt 0$ .

To find the total time of travel $T$ we just add this result to $t$ to find that

$T = (2 + \sqrt{2})t$ , giving us $K = 2 + \sqrt{2} = \boxed{3.414}$ to $3$ decimal places.