Question

The 1800-kg car descends a hill of angle 0 3°. The drag force due to air resistance is k 0.45 kg/m D kv2 for Use the sketch to complete a free-body diagram. From an initial velocity of 29 m/s (65 mph), the car coasts 1600 m along the slope. Determine the velocity at that point Hint: After finding the acceleration, choose a dr vs dvs to integrate the kinematics; the resulting integral simplifies with the substitution, u = denominator, which leads to a du/u form A: v 39 m/s

Answer 1

From Newton�s second law of molar F = mg sin theta - D ma = mg sin theta - R v^2 mvdv/dx = mg sin theta - R v^2 v dv/(g sin theta - R/m v^2) = dx Let g sin theta = a R/m = b v dv/(a - bv^2) = dx therefore (- 1/2b) d(a - bv^2)/(a - bv^2) = dx {d (a - bv^2) = -2bvdv d(a - bv^2)/(- 2b) = vdv theta r = d(a - bv^2)/(a - bv^2) = -2bdx integrating from v_initial = 29 m/s x_i = 0 v_final = v_f x_f = 1600m integral d(a - bv^2)/(a - bv^2) = -2b integral dx ln (a - bv^2) Vertical-bar ^v_f_v_i = -2b (x_f) ln [a - bv^2_f/a - bv^2_i] = -2bx_f therefore a - bv^2_f = (a - bv^2_i) e^- 2bx_f \r\n a - bv^2_f = (a - bv^2_i) e^- 2bx_f - bv^2_f = -a + (a - bv^2_i) e^- 2bx_f v_f = [a/b - (a/b - v^2_i) e^- 2bx_f]^1/2 because a = g sin theta = 9.8 times sin 3 degree