A particle starts from rest and travels with constant acceleration ‘$a$’. After a time of '$t_0$' its acceleration direction gets reversed.  Identify correct options.

The particle will travel for a time of t₀, so the velocity at that instant will be at₀.

And the particle will travel a further for a time of t₀ and come to rest, the distance travelled will be the same distance where it was accelerated.

So the distance at which it comes to rest is $2\times \frac{1}{2}\times \mathrm{a}\times {\mathrm{t}}_0^2={\mathrm{at}}_0^2$.

While travelling back it will be accelerated in the opposite direction from rest for a distance of $a{t}_0^2$, so the total distance travelled will be $2a{t}_0^2$.

Time taken for return will be given by $a{t}_0^2=\frac{1}{2}a{T}^2\Rightarrow T=\sqrt{2}{t}_0$

So the total time of travel will be $\left(2+\sqrt{2}\right)t_0$.

The speed at which it reaches the initial position will be $v=aT=\sqrt{2}a{t}_0$.