Q132 / 180 · AIIMS 2008 · Physics

The displacement time graph of a particle moving along a straight line is drawn below. Figure

  • A
    \(\begin{array}{c|c} \text{OP} & \text{PQ} & \text{QR} \\ \hline - & 0 & + \end{array}\)
  • B
    \(\begin{array}{c|c} \text{OP} & \text{PQ} & \text{QR} \\ \hline + & 0 & +\end{array}\)
  • Correct answer
    C
    \(\begin{array}{c|c} \text{OP} & \text{PQ} & \text{QR} \\ \hline + & 0 & -\end{array}\)
  • D
    \(\begin{array}{c|c} \text{OP} & \text{PQ} & \text{QR} \\ \hline - & 0 & -\end{array}\)

Solution

Acceleration $=\frac{d^2 x}{d t^2}$ For part $O P$, the acceleration is positive as the velocity is increasing. Slope is increasing. For part $P Q$, the acceleration is zero. Horizontal straight line shows that the velocity is zero. For part $Q R$, acceleration is negative. It is retarded motion. Velocity is decreasing as the slope decreases with time. Hence option (c) is correct.

Hence, the correct answer is option: 'C'.

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