The horizontal coordinates of a Frisbee in a strong wind are given by x = -12 t

Question

The horizontal coordinates of a Frisbee in a strong wind are given by x = -12 t +4t^2 and y = 10 t -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. (b) Give a direction, measuring angles from the positive x direction (c) What is the magnitude of the velocity at t = 2.0 s, accurate to the nearest m/s?

Explanation / Answer

vx = -12 + 8t ax = 8 vy = 10-6t ay = -6 >br/> a ) acceleration = sqrt [ 8^2 + (-6)^2] = 10,/s^2 b ) tan {theta} = -6 / 8 = {theta} = -36.86 degree .............which can be expressed as...36.86 degree from X axis..in clockwise direction.......or 323.14 degrees measure anticlockwise... c) at t = 2 s, vx = -12+16 = 4m/s ; vy = 10 - 12 = -2 m/s so, v = sqrt ( 4^2 + 2^2) = 4.472 m/s

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