A particle moves in the x-y plane with a y-component of velocity in feet per second given by vy = 5.5t with t in seconds. The accelerationof the particle in the x-direction in feet per second squared is given by ax = 3.3t with t in seconds. When t = 0, y = 3.5 ft, x = 0, and vx = 0.The equation of the path of the particle can be written in the form (y - b)³ = cx². Find the constants b and c and calculate the magnitudeof the velocity v of the particle for the instant when its x-coordinate reaches 17.4 ft.Answer:b=C =iiWhen x = 17.4 ft, v =iftftft/sec

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