Question

The horizontal coordinates of a frisbee in a strong wind are given by x = -12t + 4t^2 and y = 10t - 3t^2, where x and y are in meters, and t is in seconds.

(a) What is the acceleration of the frisbee? Give a magnitude and a direction, measuring angles from the positive x direction.
(b) In unit vector notation, what is the velocity of the frisbee at its greatest y coordinate?

Answer 1

r = -12t + 4t^2 y = 10t - 3t^2 Acceleration (a^vector) = d^2x/dt^2 + d^2y/dt^2 d^2x/dt^2 = d/dt (- 12 + 8t) a_x = (- 0 + 8) = 8 m/s^2 d^2y/dt^2 = d/dt (10 - 6t) a_y = (0 - 6) m/s^2 Therefore a^vector = [8 + (- 6)) m/s^2 magnitude (a^vector) = squareroot 8^2 + 6^2 = 10 m/s^2 direction Theta = Tan^- 1(6/8) = 36.87 degree from positive X - axis below = 36.87 degree velocity v^vector = v_2 + v_y V^vector = (- 12 + 8t) + (10 - 6t) at greatest y - e coordinate, v_y = 0 10 - 6t = 0 t = 10/6 = 1.67 sec Therefore v^vector = -12 + 8(1.67) V^vector = + 1.33 m/s Therefore V_x = v_ox + a_xt v_x = 1.33 + (10) times 1.67 find using (v_x) = 18.03 m/s Find velocity at greatest y-coordinate v = 18.03 m/s