Question

The graph between the displacement $x$ and time $t$ for a particle moving in a straight line is shown in figure. During the interval $OA,AB,BC$ and $CD$, the acceleration of the particle is

• A. $\begin{array}{cccc}OA& AB& BC& CD\\ +& 0& +& +\end{array}$
• B. $\begin{array}{cccc}OA& AB& BC& CD\\ -& 0& +& 0\end{array}$
• C. $\begin{array}{cccc}OA& AB& BC& CD\\ +& 0& -& +\end{array}$
• D. $\begin{array}{cccc}OA& AB& BC& CD\\ -& 0& -& 0\end{array}$

Answer 1

The correct option is B $\begin{array}{cccc}OA& AB& BC& CD\\ -& 0& +& 0\end{array}$

In portion $OA$, slope of $x-t$ graph is decreasing
$\therefore$ Velocity is decreasing
$\Rightarrow$ Acceleration is negative.
In portion $AB$, displacement is constant due to which $v=0$
$\Rightarrow$Acceleration$=0$.
In portion $BC$, slope of $x-t$ graph is increasing
$\therefore$ Velocity is increasing
$\Rightarrow$ Acceleration is positive.
In portion $CD$, slope is 0 due to which Velocity, $v=0$
$\Rightarrow$ Acceleration $=0$