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JEE Main 2018 (15 Apr, Shift 2) · Relative Velocity in Two Dimensions

Relative Velocity in Two Dimensions question from JEE Main 2018 (15 Apr, Shift 2)

Q22JEE Main 2018 · 15 APR · SHIFT 2Difficult

Relative Velocity in Two Dimensions

A man in a car at location Q on a straight highway is moving with speed vv. He decides to reach a point P in a field at a distance d from the highway (point M) as shown in the figure. Speed of the car in the field is half to that on the highway. What should be the distance RM, so that the time taken to reach P is minimum?

Figure

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Answer: (d)

  1. Let RM=xRM = x. Time T=QM−xv+x2+d2v/2T = \frac{QM - x}{v} + \frac{\sqrt{x^2 + d^2}}{v/2}
  2. dTdx=−1v+2xvx2+d2=0⇒2x=x2+d2\frac{dT}{dx} = -\frac{1}{v} + \frac{2x}{v\sqrt{x^2 + d^2}} = 0 \Rightarrow 2x = \sqrt{x^2 + d^2}
  3. 4x2=x2+d2⇒x=d34x^2 = x^2 + d^2 \Rightarrow x = \frac{d}{\sqrt{3}}
  4. So the answer is (d).
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