Rain is falling at a speed of 40 m s ⁻¹ at an angle of 30^ with the vertical in the forward direction. A person starting from rest accelerates uniformly at 2 m s ⁻² along a horizontal road in the same plane. The distance covered by the person at the exact moment the rain appears to fall vertically is

Step-by-step solution

Let the actual velocity of the rain be v _r . Its horizontal and vertical components are: v_ rx = 40 (30^ ) = 40 1 2 = 20 m s ⁻¹ v_ ry = 40 (30^ ) = 20 3 m s ⁻¹ For the rain to appear falling vertically to the person, the horizontal velocity of the person must be equal to the horizontal component of the rain's velocity. Thus, the required velocity of the person is v = 20 m s ⁻¹ . The person starts from rest ( u = 0 ) and accelerates at a = 2 m s ⁻² . Using the third equation of motion to find the distance s : v^2 =