Question

3) At the instant shown, car A has a constant velocity of 90 km/h, and car B has a speed of 108 km/h and is gaining speed at the rate of 5m/s2. Find the corresponding velocity and acceleration of car B with respect to car A. BOX IN YOUR ANSWERS 20° 37D

Answer 1

V_A = 90 km/hl = 90 times 1000/3600 = 25 m/s V_B = 108 km/hl = 108 times 1000/1500 = 30 m/s Now frn cas A, the derivation of velocity is direction = (- cos 20 degree, -sin 20 degree) vector V_A = 25(- cos 20 degree, -sin 20 degree) for B, the direction of velocity is direction = (cos 53, -sin 53) vECTOR v_B = (cos 53, -sin 53) Now Vector V_BA = vector V_B - vector V_A = 30(cos 53, -sin 53) - 25 (-cos 20, -sin 20) vector V_BA = (41.5467, -15.40856) now the normal acclearation of B V^2/r = 30 times 30/500 = 1.8 m/s^2 direction (-cos 37, -sin 37) vector O_Bn = 1.8 (-cos 37, -sin 37) magnitude of tangential acceleration = 5 m/s^2 vector a_b = vector BN + vector a_bf vector _BA = vector a_BA - vector a_A = 0 vector a_BA = vector a_B vector a_BA = 1.8(-cos 37, -sin37) + 5(cos 53, -sin 53) vector a_BA = (1.5715, -5.07644)