Vector Algebra
Relative Velocity
Grade 12
Question:
<p>Two particles start simultaneously from the same point and move along two straight lines, one with uniform velocity <span>\(\vec{u}\)</span> and the other from rest with uniform acceleration <span>\(\vec{f}\)</span>. Let <span>\(\alpha\)</span> be the angle between their directions of motion. The relative velocity of the second particle with respect to the first is least after a time</p>
<p>\(\dfrac{u \sin\alpha}{f}\)</p>
<p>\(\dfrac{f \sin\alpha}{u}\)</p>
<p>\(u \sin\alpha\)</p>
<p>\(\dfrac{u \cos\alpha}{f}\)</p>
Step-by-Step Solution
Key Concept: The relative velocity is minimized when the relative position vector is perpendicular to the relative velocity vector. This occurs when d|v_rel|/dt = 0, which gives the condition for minimum relative speed.
Step 1: Set up position and velocity vectors. Particle 1: v_1 = u (constant) Particle 2: v_2 = f t (starts from rest with acceleration f ) Step 2: Find relative velocity. v_rel = v_2 - v_1 = f t - u Step 3: Minimize | v_rel |^2. | v_rel |^2 = | f t - u |^2 = f^2t^2 - 2t( f · u ) + u^2 Step 4: Take derivative and set to zero. d/dt(| v_rel |^2) = 2f^2t - 2( f · u ) = 0 Since f · u = fu cos α: Step 5: Solve for time. 2f^2t = 2fu cos α t = (u cos α)/f ∴ Answer: D
Correct Answer: D