Vector Algebra
Relative Velocity using Vectors
Grade 12
Question:
<p>Two particles start from the same point. The 1st particle moves with uniform velocity \(u\) and the 2nd particle starts from rest with uniform acceleration \(f\). The angle between their directions of motion is \(\alpha\). The relative velocity of the 2nd particle with respect to 1st is \(R = \sqrt{f^2t^2 + 4^2 - 2ftu\cos\alpha}\). For the least value of \(R\) (relative velocity), \(\dfrac{dR}{dt} = 0\). The time at which the relative velocity is minimum is</p>
<p>\(t = \dfrac{u\sin\alpha}{f}\)</p>
<p>\(t = \dfrac{u\cos\alpha}{f}\)</p>
<p>\(t = \dfrac{f}{u\cos\alpha}\)</p>
<p>\(t = \dfrac{u}{f\cos\alpha}\)</p>
Step-by-Step Solution
Key Concept: The relative velocity magnitude R is minimized when dR/dt = 0. Since R² = f²t² + u² - 2ftu cos α, differentiating R² with respect to t and setting d(R²)/dt = 0 is computationally simpler than differentiating R directly, yielding t = (u cos α)/f.
Step 1: Write the relative velocity squared to simplify differentiation: R^2 = f^2t^2 + u^2 - 2ftu cos α Step 2: Differentiate R^2 with respect to t: d(R^2)/dt = 2f^2t - 2fu cos α Step 3: For minimum R, set dR/dt = 0, which means d(R^2)/dt = 0 (since 2R · dR/dt = 0): 2f^2t - 2fu cos α = 0 Step 4: Solve for t: 2f^2t = 2fu cos α t = (u cos α)/f ∴ Answer: B
Correct Answer: B