The graph between the displacement $x$ and time $t$ for a particle moving in a straight line is shown in the f

Question

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

Accepted Answer

1. **Curvature and Acceleration:** In a displacement-time ($x-t$) graph, the acceleration corresponds to the curvature of the graph (the second derivative $\frac{d^2x}{dt^2}$).
2. **Sign Convention:** 
   - **Concave Upwards:** If the graph curves upwards (like a cup), the slope (velocity) is increasing, indicating **positive acceleration** ($a > 0$) .
   - **Concave Downwards:** If the graph curves downwards (like an inverted cup), the slope (velocity) is decreasing, indicating **negative acceleration** ($a < 0$) .
   - **Straight Line:** If the graph is a straight line, the slope is constant (constant velocity), indicating **zero acceleration** ($a = 0$).
3. **Analysis of Intervals (based on the correct option):**
   - **OA:** The graph is concave downwards ($-$).
   - **AB:** The graph is a straight line ($0$).
   - **BC:** The graph is concave upwards ($+$).
   - **CD:** The graph is a straight line ($0$).
4. **Conclusion:** The sequence of acceleration signs is – 0 + 0.

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