Question

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

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

• A. A
• B. B
• C. C
• D. D

Hint:

If \(v\) vs \(x\) is a straight line, \(a\) vs \(x\) must also be a straight line because \(a = v \times (\text{constant slope})\).

Answer 1

The Correct Option is C

Step 1: From the graph, \(v\) is a linear function of \(x\): \(v = -\left(\frac{v_0}{x_0}\right)x + v_0\).
Step 2: Acceleration is given by \(a = v\frac{dv}{dx}\).
Step 3: Calculate the slope \(\frac{dv}{dx} = -\frac{v_0}{x_0}\) (a constant).
Step 4: Multiply by \(v\): \[a = \left[ -\left(\frac{v_0}{x_0}\right)x + v_0 \right] \times \left(-\frac{v_0}{x_0}\right) = \left(\frac{v_0}{x_0}\right)^2 x - \frac{v_0^2}{x_0}\]
Step 5: This is an equation of a straight line \(a = mx + c\) where the slope \(m = (v_0/x_0)^2\) is positive and the intercept \(c = -v_0^2/x_0\) is negative. Thus, it is an increasing line starting from a negative value.