A particle of mass $m$  moves on the $x$ ‐axis as follows: it starts from rest at $t=0$  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 times $(0<t<1)$ . If  $\alpha$  denotes the instantaneous acceleration of the particle, then,

The positions of the particle at $t=0$  and at   $t=1\text{s}$ are
  $x\left(0\right)=0\text{m},$$x\left(1\right)=1\text{m}.$
The particle is at rest at   $t=0$ and comes to rest at $t=1\text{s}$ . Thus, initial and final velocities of the particle are
 $v\left(0\right)=0,$$v\left(1\right)=0$ .   (1)
The velocity $v\left(1\right)=0$  of a particle having instantaneous acceleration  $\alpha$  is given by
$v\left(1\right)=v\left(0\right)+\underset{0}{\overset{1}{ }}\alpha \text{d}t$.   (2)

The equations (1) and (2) give,$\underset{0}{\overset{1}{ }}\alpha \text{d}t=0$  . Thus, area under the $\alpha -t$  curve is zero. If $\alpha =0$  for all  $0<t<1$  then area becomes zero but this also makes  $x\left(1\right)=0,$  which is not true. If  $\alpha >0$  for all  $0<t<1$ then
area under the curve is positive and if  $\alpha <0$ for all   $0<t<1$ then area under the curve is negative. Thus,  $\alpha$  must change sign at some time in $0<t<1$  . Only conclusions that can be made are $\underset{0}{\overset{1}{ }}\alpha \text{d}t=0,\alpha \ne 0,$  and $\alpha$   must change sign (at least once) in  $0<t<1.$

If $\alpha$ change sign only once (say at  $t=t_c$) and  $\alpha ={\alpha }_1$ in $0<t<t_c$  and $\alpha =-{\alpha }_2$   in $t_c<t<1$ , where ${\alpha }_1$  and ${\alpha }_2$   are some positive constants, then condition $x\left(1\right)=1\text{m}$   is violated if $\alpha <4$  at all points. We encourage you to explore area under v‐t graph to show that $\alpha >4$  at some point(s). Hint: The area under v‐t graph, $\frac{1}{2}\times 1\times v_{\text{ max }}=1$ , gives $v_{\text{ max }}=2\text{m}/\text{s}.$  Now,  $v_{\text{ max }}=2={\alpha }_1{t}_c={\alpha }_2(1-t_c)$ . If $t_c<\frac{1}{2}$  than   ${\alpha }_1>4$ otherwise  ${\alpha }_2>4.$