Applications of Derivatives
Kinematics / Rate of change
Grade 12
Question:
<p>If the law of linear motion of a particle is given by \(s = \dfrac{1}{3}t^3 - 16t\), then the acceleration at the time when the velocity vanishes, is</p>
<p>(a) 0</p>
<p>(b) 4</p>
<p>(c) 8</p>
<p>(d) none of these</p>
Step-by-Step Solution
Key Concept: Find when velocity equals zero by differentiating position, then evaluate acceleration (second derivative) at that time point.
<p><strong>Step 1:</strong> Find velocity by differentiating position.</p><p>Given: s = (1/3)t³ - 16t</p><p>v = ds/dt = t² - 16</p><p><strong>Step 2:</strong> Find when velocity vanishes (v = 0).</p><p>t² - 16 = 0</p><p>t² = 16</p><p>t = ±4</p><p>Since time is physical, take t = 4 (or both are valid for acceleration)</p><p><strong>Step 3:</strong> Find acceleration by differentiating velocity.</p><p>a = dv/dt = 2t</p><p><strong>Step 4:</strong> Evaluate acceleration at t = 4.</p><p>a = 2(4) = 8 m/s²</p><p>∴ Answer: C (acceleration = 8 m/s²)</p>
Correct Answer: C