Question

A particle moves along $x$-axis. it is at rest, at $t=0,x=0$ and comes to rest at $t=1,x=1$. If $\alpha$ denotes instantaneous acceleration, then

• A. $\alpha$ cannot remain positive in the interval $0\le t\le 1$
• B. $\left|\alpha \right|\le 2$ at any point in the path
• C. $\left|\alpha \right|$ must be $\ge 4$ at some point in the path
• D. $\alpha$ must change sign during motion but no other assertion can be made

Answer 1

Answer: (A), (C)

The correct options are
A $\alpha$ cannot remain positive in the interval $0\le t\le 1$
C $\left|\alpha \right|$ must be $\ge 4$ at some point in the path

If $\alpha$ is positive in $0\le t\le 1$ then it will not stop at $t=1$ because there has to be some deceleration to stop the particle.
At t=0.5 we will have x=0.5 by symmetry. Thus using $s=ut+\frac{1}{2}a{t}^2$ we get
$0.5=\left(0\right)t+\frac{1}{2}a(0.5{)}^2$ we get a as $4m/s^2$.
Thus C is correct.