Question

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 a denotes the instantaneous acceleration of the particle, then

• A. $\alpha$ cannot remain positive for all $t$ in the interval 0 $\le t\le 1$
• B. $|\alpha |$ cannot exceed 2 at any point in its path
• C. $|\alpha |$ must be $\ge$ 4 at some point or points in its path
• D. $\alpha$ must change sign during the motion, but no other
  assertion can be made with the information given

Answer 1

Answer: (A)

Since, the body is at rest at $x=0$ and $x=1$ Hence, a cannot
be positive for all time in the interval 0$\le t\le 1$
Therefore, first the particle is accelerated and then retarded.
Now, total time $t=1s$ (given)
Total displacement $s=1m$ (given)
$s$ = Area under $v-t$ graph
$\therefore$ Hight or $v_{max}=\frac{2s}{t}=2m/s$ is also fixed
[Aera or $s=\frac{1}{2}\times t\times v_{max}$]
If height and base are fixed, area is also fixed .
In case 2 : Acceleration = Retardation $=4m/s^2$
In case 1 : Acceleration $>4m/s^2$ while
Retardation $<4m/s^2$
While in case 3 : Acceleration $<4m/s^2$ as
Retardation $>4m/s^2$
Hence,$|\alpha |\ge 4$ at some point or points in its path.