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The particle travels along the path defined by the parabola y=0.5x2, where x and y are in ft. 1- If the component of velocity along the x axis is vx=(8t)ft/s, where t is in seconds, determine the particle's distance from the origin O t = 3 s. When t=0, x=0, y=0. 2- Determine the magnitude of its acceleration when t = 3 s.

Respuesta :

Answer:

Explanation: dx/dt=5t (integrate) x=5/2t^2+c (plug in initial condition to solve for c) 0=5/2(0)^2+c ----> c=0 x=5/2t^2 (plug in 1) ---> x = 5/2(1)^2 = 5/2 y=25/8t^4 (plug in 1) ---> x = 25/8(1)^4 = 25/8 solve for distance sqrt((5/2)^2+(25/8)^2) sqrt(25/…

Answer:

288 ft

8 ft²/s

Explanation:

We can determine the particles distance from the origin by substitute t into the velocity equation and then in the path equation:

For t=3:

vx [tex]=8\cdot{3}=24[/tex]

[tex]y=0.5\cdot{24^2}=288[/tex]

The particle travels 288 ft

The acceleration is the derivative of the velocity:

[tex]a=8[/tex]

8 ft²/s and will stay constant with time.

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