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October 30, 2014

October 30, 2014

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

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

- Engineering Kinematic -
**drwls**, Saturday, October 16, 2010 at 6:29amWhen 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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