The velocity - displacement graph of a particle is shown in the figure.
The acceleration - displacement graph of the same particle is represented by :

Solution

Physics sir · 4 steps

On a v–x graph the slope is not acceleration: multiply it by v, since a = v dv/dx.

1. Equation of the line. The line passes through (0, v₀) and (x₀, 0), so v = v₀(1 - x/x₀). Its slope is dv/dx = -v₀/x₀, a negative constant.
2. Chain rule. By the chain rule, a = dv/dt = dv/dx·dx/dt = v dv/dx. Put in both pieces.

   a = v₀(1 - x/x₀)(-v₀/x₀) = -v₀²/x₀ + v₀²/x₀² x

3. Read the shape. This is a straight line in x with a positive slope v₀²/x₀². At x = 0, a = -v₀²/x₀ (negative), and at x = x₀, a = 0.
4. So. The a–x graph starts at a negative value on the a-axis and rises linearly to zero at x₀. That is option (a).

The idea to keep. Whenever velocity is given against position, use a = v dv/dx. A constant slope on a v–x graph does not mean constant acceleration, because v itself keeps changing.

Answer from the NTA final answer key. Solution by PrepWiser.