Figure shows two rods p & q moving with a same velocity of 2 m/s perpendicular to their length. If angle between the rods is 106∘, velocity of point of intersection (in m/s) is.
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Solution
Say, DB = CB = x By triangle ACB, sinθ=BCAB ⇒AB=BCsinθ By differentiating the equation on both sides, vA=vsinθ=2sin53=2×54=2.5m/s