Acceleration Dependence on Position from Equation

❓ Question

The distance $x$ covered by a particle in one dimensional motion varies with time $t$ as:

$$x^2=at^2+2bt+c$$

If the acceleration of the particle depends on:

$$x^{-n}$$

where $n$ is an integer, then the value of $n$ is ______.

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🖼 Question Image

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✍️ Short Explanation

This problem connects:

👉 Displacement-time relation
👉 Velocity and acceleration derivation
👉 Dependence of acceleration on displacement.

Main goal:

$$a \propto x^{-n}$$

Find value of $n$.

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🔷 Step 1 — Given Equation 💯

Given:

$$x^2=at^2+2bt+c$$

Differentiate with respect to time:

$$2x\frac{dx}{dt}=2at+2b$$

Since:

$$\frac{dx}{dt}=v$$

So:

$$xv=at+b$$
$$v=\frac{at+b}{x}$$

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🔷 Step 2 — Find Velocity Relation

Square both sides:

$$v^2=\frac{(at+b)^2}{x^2}$$

From original equation:

$$x^2=at^2+2bt+c$$

Now:

$$(at+b)^2=a(at^2+2bt+c)+(b^2-ac)$$

Thus:

$$(at+b)^2=ax^2+(b^2-ac)$$

Therefore:

$$v^2=a+\frac{b^2-ac}{x^2}$$

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🔷 Step 3 — Use Acceleration Formula

Using:

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

Differentiate:

$$v^2=a+\frac{k}{x^2}$$

where:

$$k=b^2-ac$$

Differentiate w.r.t $x$:

$$2v\frac{dv}{dx}=-\frac{2k}{x^3}$$

So:

$$v\frac{dv}{dx}=-\frac{k}{x^3}$$

Hence acceleration:

$$\boxed{a\propto x^{-3}}$$

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🔷 Step 4 — Compare with Given Form

Given:

$$a\propto x^{-n}$$

Comparing:

$$x^{-n}=x^{-3}$$

Therefore:

$$\boxed{n=3}$$

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🔷 Step 5 — JEE Trap Alert 🚨

❌ Direct second differentiation kar dena

❌ $v=a/t$ assume kar lena

❌ Chain rule:

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

bhool jaana

Remember:

Whenever $v^2$ is function of $x$,

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

is fastest method.

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✅ Final Answer

$$\boxed{3}$$

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📚 Related Topics

• Relative Motion with Upward Accelerating Frame
• Displacement of Charged Particle in Two B Fields
• Acceleration Velocity Displacement vs Time Graphs