At a particular instant, two cars A and B are traveling at 20 km/hr and 40 km/hr. Car B is decreasing its speed by 1600 km/hr2 while A is decreasing its speed at 1000 km/hr2. A ball thrown by a person in A is caught by a person in car B. Determine the velocity of the ball. Also find the acceleration of B with respect to A, if B moves along a curve having radius of curvature as 2 kms. 1) 29.46 km/hr & 2748.8 km/hr2 2) 22.34 km/hr & 4567.4 km/hr2 3) 32.78 km/hr & 1234.3 km/hr2 4) 87.45 km/hr & 1245.6 km/hr2. The fixed axes x,y are established at a point on the ground,while translating x',y' axes are attached to car A. Velocity The velocity of the ball is the velocity of the car B with respect to car A. VB = VA + VB/A. 40 sin45 i + 40 cos45 j = 20 i + VB/A VB/A = 8.284 i + 28.284 j = 29.46 km/hr. The angle made by the VB/A to positive x axis is given by q= arctan(28.28/8.28) = 73.68o Acceleration The accleration of B along the radius of curvature is an = (VB2)/r = 1600/2 = 800 km/hr2. The tangential and normal accelerations, have to be considered by taking their components along x,y axes. aB = -800cos45 i + 800 sin45 j - 1600 cos45 i - 1600 cos45 j = 1690.056 i - 565.68 j aA = -1000i aB = aA + aB/A aB/A = 2690.056 i - 565.68 j = 2748.8 km/hr2 The angle made by this vector with the positive x axis is given by q= arctan(-565.68/2690.056) = -11.875o