Question

The displacement versus time curve is given below. Match the nearest option. (Assuming curves to be parabolic)


$\begin{array}{cc}\text{Column I}& \text{Column II}\\ i.OA& \text{a. velocity increase with time linearly}\\ ii.AB& \text{b. velocity decreases with time}\\ iii.BC& \text{c. velocity is independent of time}\\ iv.CD& \text{d. velocity is zero}\end{array}$

• A. $i.\to a.,ii.\to c.,iii.\to b.,iv.\to d$.
• B. $i.\to a.,ii.\to c.,iii.\to d.,iv.\to b$.
• C. $i.\to c.,ii.\to a.,iii.\to b.,iv.\to d$.
• D. $i.\to a.,ii.\to b.,iii.\to c.,iv.\to d$.

Answer 1

The correct option is A $i.\to a.,ii.\to c.,iii.\to b.,iv.\to d$.

$i.\to a.,ii.\to c.,iii.\to b.,iv.\to d$.
In OA, $S\propto t^2,v=\frac{dS}{dt}\propto 2t,i.e.,v\propto t.$
i.e., velocity increases with time.
In AB, $S\propto t,v=\frac{dS}{dt}\propto t^0$
i,e., velocity is uniform, i.e., constant or independent of time.
In BC, body is retarded, i.e., velocity decreases with time.
In CD, $S\propto t^0$ i.e., $v$ = zero i.e., body is at rest.