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Posted by on Saturday, October 16, 2010 at 4:49am.

The particle travels along the path defined by the parabola y=0.5(x^2). If the component of velocity along the x-axis is v=5t [m/s], determine the particle's distance from the origin and the magnitude of its acceleration when t=1s. The initial condition is t=0, x=0 and y=0.

Can someone please tell me tha answers and how to work it please?

  • Engineering Kinematic - , Saturday, October 16, 2010 at 6:29am

    When t = 1, dx/dt = 5t = 5 m/s
    From v = dx/dt = 5t, and the starting conditions, you can conclude that
    x = (5/2)t^2.
    x = 5/2 = 2.5 when t = 1
    y = 0.5*(2.5)^2 = 3.125 when t = 1
    The distance from the origin in sqrt[x^2 + y^2]. At t=1, this is
    D = sqrt[2.5^2 + 3.125^2]= 4.002

    The acceleration components are
    a_x = d^2x/dt^2 = dv_x/dt = 5 @t=1)
    and
    a_y = d^2y/dt^2 = dv_y/dt
    v_y = dy/dt = dy/dx*dx/dt = x*v_x =
    a_y= dv_y/dt = (v_x)^2 + a_x*x
    = 25 + 5*1 = 30

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