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) Write the relation between $v$ and $x$. (b) Obtain the relation between acceleration and displacement and plot it.

The velocity–displacement graph of a particle is a straight line: the velocity has its maximum value v₀ on the

NCERT Exemplar · Q22

v_0

(a) Write the relation between

v

and

x

(b) Obtain the relation between acceleration and displacement and plot it.

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Concept understanding — Instantaneous Rate Of Change

Instantaneous Rate of Change

Imagine you're in a car watching the speedometer. It doesn't say "I travelled 60 km in the last hour" — it shows your speed right now, at this exact moment. That number, the one that changes every time you tap the brake or press the accelerator, is the instantaneous rate of change of your position.

The intuition: from average to instant

If you drive from Delhi to Agra (200 km) in 4 hours, your average speed is 50 km/h. But that tells you nothing about how fast you were going at 10:15 AM when you passed a particular toll booth. You might have been doing 80 km/h, or 20 km/h if there was traffic.

The average rate of change over a time interval $[t_1, t_2]$ is:

\text{Average speed} = \frac{\text{distance travelled}}{\text{time taken}} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}

where $s(t)$ is your position at time $t$ .

To get the speed at a specific moment $t = a$ , you'd want to look at smaller and smaller intervals around $a$ . If you measure from $t = a$ to $t = a + h$ , where $h$ is a tiny time difference:

\text{Average speed over } [a, a+h] = \frac{s(a+h) - s(a)}{h}

As $h$ gets smaller — say 0.1 seconds, then 0.01, then 0.0001 — this average speed gets closer and closer to a single number. That limiting number is the instantaneous rate of change at $t = a$ .

Note

This is the core idea: instantaneous rate of change = the limit of average rates of change as the interval shrinks to zero.

The precise definition

For any function $y = f(x)$ , the instantaneous rate of change at $x = a$ is:

\text{Instantaneous rate of change} = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

provided this limit exists. This limit is also called the derivative of $f$ at $a$ , denoted $f'(a)$ or $\frac{dy}{dx}\big|_{x=a}$ .

f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

What it means geometrically

If you graph $y = f(x)$ , the average rate of change over $[a, a+h]$ is the slope of the secant line through $(a, f(a))$ and $(a+h, f(a+h))$ . As $h \to 0$ , that secant line pivots and approaches a tangent line at $x = a$ . The slope of that tangent line is exactly $f'(a)$ .

So instantaneous rate of change = slope of the tangent line.

A concrete example

Let $f(x) = x^2$ . Find the instantaneous rate of change at $x = 3$ .

First, the average rate over $[3, 3+h]$ :

\frac{f(3+h) - f(3)}{h} = \frac{(3+h)^2 - 9}{h} = \frac{9 + 6h + h^2 - 9}{h} = \frac{6h + h^2}{h} = 6 + h

Now take the limit as $h \to 0$ :

\lim_{h \to 0} (6 + h) = 6

So at $x = 3$ , the function $x^2$ is changing at a rate of 6 units per unit change in $x$ . The tangent line at $(3, 9)$ has slope 6. …

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The graph is a line through $(0,v_0)$ and $(x_0,0)$ , so $v=v_0(1-x/x_0)$ . To get acceleration, use the chain rule $a=\dfrac{dv}{dt}=v\dfrac{dv}{dx}$ ; with the constant slope $\dfrac{dv}{dx}=-\dfrac{v_0}{x_0}$ this gives a linear $a$ – $x$ relation increasing from $-v_0^2/x_0$ at $x=0$ to $0$ at $x=x_0$ .

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

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

\frac{dv}{dx}=\frac{0-v_0}{x_0-0}=-\frac{v_0}{x_0}.

Therefore

v = v_0 - \frac{v_0}{x_0}\,x = v_0\left(1-\frac{x}{x_0}\right).

(b) Relation between acceleration and displacement

Acceleration relates to displacement through

a=\frac{dv}{dt}=\frac{dv}{dx}\cdot\frac{dx}{dt}=v\,\frac{dv}{dx}.

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

a = \left[v_0\left(1-\frac{x}{x_0}\right)\right]\left(-\frac{v_0}{x_0}\right) = -\frac{v_0^2}{x_0}\left(1-\frac{x}{x_0}\right)=\frac{v_0^2}{x_0^2}\,x-\frac{v_0^2}{x_0}.

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

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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

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

v = v_0\left(1-\frac{x}{x_0}\right)

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

\frac{dx}{dt} = v_0 - \frac{v_0}{x_0}x

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

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:

x(t) = x_0\left(1-e^{-kt}\right), \qquad k = \frac{v_0}{x_0}

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.

Differentiate once to confirm $v(t)$ :

v(t) = \frac{dx}{dt} = x_0k\,e^{-kt} = v_0\,e^{-kt} \qquad (\text{since } x_0k = x_0\cdot v_0/x_0 = v_0)

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

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

a(t) = \frac{dv}{dt} = -v_0k\,e^{-kt} = -k\,v(t)

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: …

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