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JEE PYQ: Motion in a Straight Line - Question ID b707b9d06c61 (JEE Main 2021)

ID: b707b9d06c61JEE Main 2021Numerical Value

A particle is moving with constant acceleration 'a'. Following graph shows v² versus x(displacement) plot. The acceleration of the particle is ___________ m/s².

JEE Main 2021 (Online) 31st August Evening Shift Physics - Motion in a Straight Line Question 68 English

JEE Main 2021 (Online) 31st August Evening Shift Physics - Motion in a Straight Line Question 68 English

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Step-by-step Explanation

Core Formula & Concept:

When a particle moves with constant acceleration $a$ , its velocity $v$ and displacement $x$ are related by the kinematic identity $v^2 = v_0^2 + 2\,a\,x,$ where $v_0$ is the initial velocity at $x=0$ . This equation shows that $v^2$ is a linear function of $x$ with slope $2a$ .

Step-by-Step Derivation:

1. From the graph we see two points on the $v^2$ – $x$ line:
At $x=0$ , $v^2 = 4\;(\text{m}^2/\text{s}^2)$ .
At $x=2\;\text{m}$ , $v^2 = 8\;(\text{m}^2/\text{s}^2)$ . 2. Compute the slope $m$ of the line: $m = \frac{\Delta(v^2)}{\Delta x} = \frac{8 - 4}{2 - 0} = \frac{4}{2} = 2\;\bigl(\text{m}/\text{s}^2\bigr).$ 3. But from the kinematic relation $v^2 = v_0^2 + 2a\,x$ , the slope is $2a$ . Hence $2a = 2 \quad\Longrightarrow\quad a = 1\;\text{m/s}^2.$

Common Traps & Exam Tip:

• Students often confuse the slope of $v^2$ vs $x$ with the acceleration itself. Remember the slope is $2a$ , not $a$ .
• Always check the intercept to confirm $v_0^2$ ; here it is $4$ , so $v_0=2$ m/s, but that value is not needed for $a$ .

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