Find Time When Velocity Returns Using Acceleration Graph

❓ Question

The acceleration-time graph of a particle is as shown.

At what time does the particle acquire its initial velocity?

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🖼 Question Image

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✍️ Short Explanation

This is an acceleration-time graph problem.

👉 Change in velocity equals area under $a$-$t$ graph
👉 Particle regains initial velocity when net area becomes zero.

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🔷 Step 1 — Use Area Concept 💯

From graph:

Initial acceleration:

$$10\text{ m/s}^2$$

Acceleration becomes zero at:

$$t=4\text{ s}$$

So positive triangular area from:

$$0 \to 4$$

is:

$$A_1=\frac12 \times4\times10$$
$$=20$$

This represents increase in velocity.

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🔷 Step 2 — Condition for Initial Velocity

Particle acquires initial velocity again when:

$$\Delta v=0$$

So negative area after:

$$t=4$$

must equal:

$$20$$

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🔷 Step 3 — Equation of Straight Line

Acceleration graph is straight line.

Slope:

$$=\frac{0-10}{4-0}$$
$$=-\frac{10}{4} =-2.5$$

Equation:

$$a=10-2.5t$$

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🔷 Step 4 — Find Time for Equal Negative Area

At time $t$:

Magnitude of negative acceleration:

$$|a|=2.5t-10$$

Negative triangular area from:

$$4 \to t$$

is:

$$A_2=\frac12 (t-4)(2.5t-10)$$

Set:

$$A_2=20$$
$$\frac12 (t-4)(2.5t-10)=20$$

Since:

$$2.5t-10=2.5(t-4)$$
$$\frac12 (t-4)\cdot2.5(t-4)=20$$
$$1.25(t-4)^2=20$$
$$(t-4)^2=16$$
$$t-4=4$$
$$t=8\text{ s}$$

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🔷 Step 5 — JEE Trap Alert 🚨

❌ Velocity-time graph samajh lena

❌ Negative area ka sign ignore kar dena

❌ Total area instead of net area use karna

Remember:

$$\Delta v=\text{Area under }a-t\text{ graph}$$

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✅ Final Answer

$$\boxed{8\text{ s}}$$

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• Acceleration Displacement Graph from Velocity Graph
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