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Manipal MET2013PhysicsMotion in Two Dimensions

A man crosses a 320 m wide river perpendicular to the current in 4 min . If in still water he can swim with a speed 5 / 3 times that of the current, then the speed of the current, in mm ⁻¹ in is

Options

  1. A30
  2. B40
  3. C50
  4. D60

Correct answer

D. 60

Step-by-step solution

v_r^2=v_m^2-v^2v= 320 4 ~m / min =80 ~m / min aligned & v_m= 5 3 v_r & v_r^2=t 5 3 (v_r )^2-(80)^2 aligned 16 9 v_r^2=(80)^2u^2, u₀ form canours linear

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