A particle moves in the x-y plane with a y-component of velocity…
A particle moves in the x-y plane with a y-component of velocity in feet per second given by vy = 8t with t in seconds. The acceleration of the particle in the x-direction in feet per second squared is given by ax = 4t with t in seconds. When t = 0, y = 2 ft, x = 0, and vx = 0. Find the equation of the path of the particle and calculate the magnitude of the velocity v of the particle for the instant when its x-coordinate reaches 18 ft.
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Technifi Expert’s Answer:
Given: vy=8t which means, dy/dt=8t
y=4t^2+2 -> (1)
x=2/3 t^3 + c
substitute value of t from eq 2 in eq 1
y=4(3*x/2)^(2/3)+2 -> eq of motion
then at x=18, t will be 3 from eq (2)
mag. of resultant vector=30 ft/s
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