Question

1. (4 pts) Particles A and B start at the origin and move in the xy-plane. The velocity of Particle A is−→u=−→i−3−→j and Particle B is moving in a direction perpendicular to Particle A. After 1 unit of time, the two particles become√26 units apart and Particle B is in the first quadrant, what is the velocity of Particle B?

1. (4 pts) Particles A and B start at the origin and move in the ry-plane. The velocity of Particle A is t = 7 - 31 and Particle B is moving in a direction perpendicular to Particle A. After 1 unit of time, the two particles become 26 units apart and Particle B is in the first quadrant, what is the velocity of Particle B?

Answer 1

solution Velo city of particle A, MA = 2 - 3 velocity of particle B. Uz xity Both particles are moving perpendicular Let to each other UA. UB=0 ( 9 – 39). ( z1 + = 0 2-3y =0 => x=34 ↑ - uni Joo 29 = 2 - og B612

1 unit is hence direction particle B is in of time particle B the first quadrant first quadrant. In one unit time of Particle A will move 101)+(3) 1+=JO units particle ß will move = (x)² + (y): = I x ² + y2 units 2 of their theorem one unit of time two particles become $260 units. af art. By Pythagoras (26)*= (Iz tytu)" + + (J10 xat ty " x = 34 2 yad + 10 22 e 16 = gyot yang y = 16 4 y 10 Jio B612

DE 37 3X 4 10 0 The velocity of particle B is xit y xity 65 To ît a 10 2. To ( 3h + } UB B612