Question

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Suppose that the position of one particle at time t is given by and the position of a second particle is given by 22 3+cos(t), v2 1+sin(t), 0st s 2r. (a) How many points of intersection are there for these paths? Number of intersection points- (b) How many of these points of intersection are collision points? In other words, how many times are the particles in the same place at the same time? Number of collision points (c List the collision points. List thern in terms of increasing z coordinates. If several collision points have the same z-coordinate list hose in order of increasing y-coordinates. Type an upper-case "N" in each unused answer blank. First collision point: (r,y) -( Second collision point: (z, v)-( Third collision point: (x, y)-( (d) Now suppose that the path of the second particle is given by 2 1 + sin(t), 0 < t z2-3 + cos(t), y2

Answer 1

A) The position of one practice at time t is given by x_1 = 3 sin t y_1 = 2 cos t 0 lessthanorequalto t elementof 2 pi sin t = x_1/3 cos t = y_1/2 (Sin y)^2 + (cos t)^2 = (x_1/2)^2 + (y_1/2)^2 = x^2_1/9 + y^2_1/4 Sin^2 t + cos^2 t = x^2_1/9 + y^2_1/4 x^2_1/9 + y^2_1/4 = 1 The position of Second particle at time t is given by x_2 = -3 + cos t y_2 = 1 + sin t 0 lessthanorequalto t lessthanorequalto 2 pi cos t = x_2 + 3 sin t = y/2 = 1 (Cos t)^2 = (x_2 + 3)^2 (Sin t)^2 = (y_2 - 1)^2 cos^2 t = (x_2 + 3)^2 sin^2 t = (y_2 - 1)^2 Sin^2 t + cos^1 t = (y_2 - 1)^2 + (x_2 + 3)^2 1 = (y_2 - 1)^2 + (x_2 + 3)^2 Rightarrow (y_2 - 1)^2 + (x_2 + 3)^2 = 0 \r\n The path followed by first particle is that of an ellipse equation is x^2/9 + y^2/4 = 1