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 time $(0<t<1)$. If $\alpha$ 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 points 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), (C), (D)

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

Now $\frac{1}{2}{v}_{\max }\cdot t=s$
$\Rightarrow v_{\max }=\frac{2\times s}{t}$
$v_{\max }=\frac{2\times 1}{1}=2m/s$
The maximum velocity $=2m/s$
Now just see the $v-t$ graph
$ForABC,\left|\begin{array}{c}\text{During AB}\\ \text{During BE}\end{array}\right|\begin{array}{c}\alpha >4m/s^2\\ \alpha <-4/m{s}^2\end{array}$
$ForACE,\left|\begin{array}{c}\text{During AC}\\ \text{During CE}\end{array}\right|\begin{array}{c}\alpha =4m/s^2\\ \alpha =-4m/s^2\end{array}$
$ForADF,\left|\begin{array}{c}\text{During AD}\\ \text{During DE}\end{array}\right|\begin{array}{c}\alpha <4m/s^2\\ \alpha >-4m/s^2\end{array}$
Thus, $\alpha \ge 4$ at some point or points in its path.