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 $0 \leq t \leq 1$
• B. $|\alpha|$ cannot exceed 2 at any points in its path
• C. $|\alpha|$ must be $\geq 4$ 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 $t=1 \, s$ 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 $0 \leq t \leq 1$. If we plot $vt$ -tgraph The total displacement $=1 m =$ area under $v-t$ graph

Now $\frac{1}{2} v_{\max } \cdot t=s$
$\Rightarrow v_{\max }=\frac{2 \times s}{t}$
$v_{\max }=\frac{2 \times 1}{1}=2 \, m / s$
The maximum velocity $=2 \,m / s$
Now just see the $v-t$ graph
$For \,ABC, \begin{vmatrix}\text{During AB}\\ \text{During BE}\end{vmatrix}\begin{matrix}\alpha >4 m /s^{2}\\ \alpha < - 4 /ms^{2}\end{matrix}$
$For \,ACE, \begin{vmatrix}\text{During AC}\\ \text{During CE}\end{vmatrix}\begin{matrix}\alpha =4\, m /s^{2}\\ \alpha = -4\, m/s^{2}\end{matrix}$
$For \,ADF, \begin{vmatrix}\text{During AD}\\ \text{During DE}\end{vmatrix}\begin{matrix}\alpha < 4 \,m /s^{2}\\ \alpha > -4 \,m/s^{2}\end{matrix}$
Thus, $\alpha \geq 4$ at some point or points in its path.