A particle moves in the x-y plane with a y-component of velocity in feet per sec

Question

A particle moves in the x-y plane with a y-component of velocity in feet per second given by vy = 3.5t with t in seconds. The acceleration of the particle in the x-direction in feet per second squared is given by ax = 4.8t with t in seconds. When t = 0, y = 2.5 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 magnitude of the velocity v of the particle for the instant when its x-coordinate reaches 20.9 ft.

Explanation / Answer

b = 2.5 ft

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