Q.The velocity–displacement graph of a particle is a straight line: the velocity has its maximum value $v_0$ on the velocity axis when the displacement $x=0$, and it decreases linearly with $x$, reaching zero when $x=x_0$ (the line runs straight from the point $(0,\,v_0)$ down to the point $(x_0,\,0)$).

(a) Relation between $v$ and $x$

The line has intercept $v_0$ (at $x=0$) and slope

Therefore

(b) Relation between acceleration and displacement

Acceleration relates to displacement through

Here $\dfrac{dv}{dx}=-\dfrac{v_0}{x_0}$ (constant), so

This is linear in $x$ with a positive slope $\dfrac{v_0^2}{x_0^2}$: …

Concept: Solving the Full Time-Domain Motion, Then Eliminating $t$ — Recognising the Same Exponential Relaxation Pattern

Method: Solve the ODE $\dfrac{dx}{dt}=v_0\left(1-\dfrac{x}{x_0}\right)$ Explicitly for $x(t)$, Then Differentiate Twice (instead of using the $a=v\,dv/dx$ shortcut)

The stored answer uses the chain-rule identity $a = v\,dv/dx$ to get from the $v$-$x$ line directly to the $a$-$x$ line, without ever solving for the motion in time. This method instead does the "long way" — treats $v=dx/dt$ as a differential equation, solves it explicitly for $x(t)$, differentiates twice to get $a(t)$, and only then eliminates $t$ to recover the same $a$-versus-$x$ relation — revealing along the way that this motion has exactly the same exponential-relaxation structure as other problems in this chapter.

Steps

1. Write down the $v$-$x$ relation from the graph, as given:

2. Treat this as a differential equation for $x(t)$, since $v=dx/dt$:

This is a linear first-order ODE with a constant "driving" term $v_0$ and a decay-rate constant $k \equiv v_0/x_0$.

3. Solve it explicitly. Its equilibrium (where $dx/dt=0$) is at $x=x_0$; the general solution decaying toward that equilibrium, starting from $x(0)=0$ (the given initial position, where $v=v_0$ matches the graph's starting point), is:

This is exactly the same functional form, $x_0(1-e^{-\gamma t})$, as the exponential-relaxation motion analysed elsewhere in this chapter — recognising the $v$-$x$ line as implying this same structure is itself a useful cross-check.

4. Differentiate once to confirm $v(t)$:

Check: $v(0)=v_0$. ✓ matches the graph's starting height.

5. Differentiate a second time to get $a(t)$:

6. Eliminate $t$ to express $a$ in terms of $x$ — the actual quantity the question asks for. From Step 3, $e^{-kt} = 1-x/x_0$, so Step 4 gives $v = v_0(1-x/x_0)$ (recovering part (a), as a consistency check), and substituting this into Step 5: …