A particle of mass m moves on the x-axis as follows: it starts from rest t =0 from the point x=0 and comes to rest at t=1 at the point x=1. No other information is available about its motion at intermediate time (0<t<1). If  denotes the instantaneous acceleration of the particle, then

The body is at rest initially and again comes to rest at t =1s at position x =1.

Thus, firstly acceleration will be positive then negative. Thus  $\alpha$ have to change the direction so that body may finally come to rest in the interval $0\le t\le 1.$ If we plot v-t graph.

The total displacement =1m = area under v-t graph

Now, $\frac{1}{2}{v}_{\max }.t=s$

$\Rightarrow v_{\max }=\frac{2\times s}{t}$

$v_{\max }=\frac{2\times 1}{1}=2m/s$

The maximum velocity = 2 m/s

Now just see the v-t graph

Thus, $\alpha \ge 4$ at some point or points in its path.