The velocity-displacement graph of a particle is shown in the figure.

The acceleration-displacement graph of the same particle is represented by :

We need to find the correct acceleration-displacement ($$a-x$$) graph corresponding to the given velocity-displacement ($$v-x$$) graph of a particle.

1. Analyze the Velocity-Displacement ($$v-x$$) Graph

From the problem statement, the velocity decreases linearly with displacement $$x$$. This straight line has a positive vertical intercept and a negative slope.

We can write the equation of this line using the slope-intercept form ($$y = mx + c$$):

$$v = -mx + v_0 \quad \text{--- (Equation 1)}$$

Where:

• $$-m$$ is the negative slope of the $$v-x$$ line ($$m > 0$$).
• $$v_0$$ is the initial velocity at $$x = 0$$ (the positive vertical intercept).

2. Derive the Acceleration Expression

Acceleration $$a$$ is defined as the rate of change of velocity with respect to time ($$\frac{dv}{dt}$$). Using the chain rule, we can express it in terms of displacement $$x$$:

$$a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}$$

Since $$\frac{dx}{dt} = v$$, the relationship simplifies to:

$$a = v \cdot \frac{dv}{dx}$$

Now, let's find the components from Equation 1:

• The derivative of velocity with respect to displacement is the constant slope:

  $$\frac{dv}{dx} = -m$$

• Substitute $$v$$ and $$\frac{dv}{dx}$$ into the acceleration formula:

  $$a = (-mx + v_0) \cdot (-m)$$

  $$a = m^2x - mv_0 \quad \text{--- (Equation 2)}$$

3. Determine the Features of the $$a-x$$ Graph

Equation 2 ($$a = m^2x - mv_0$$) is also a linear equation representing a straight line on an $$a \text{ vs } x$$ coordinate plane:

• Slope: The coefficient of $$x$$ is $$m^2$$. Since any real squared number is positive, the line has a positive slope (it slants upwards from left to right).
• Vertical Intercept: At $$x = 0$$, the acceleration is $$a = -mv_0$$. Since both $$m$$ and $$v_0$$ are positive constants, the vertical intercept is negative (it starts below the horizontal origin axis).

Looking at the options, the graph that features a straight line starting from a negative vertical intercept and sloping upwards to cross the $$x$$-axis is Option C.

Conclusion

The correct acceleration-displacement graph is represented by Option C.