Answer to Question #151126 in Mechanical Engineering for Sujana

Question #151126
the motion of a particle is defined by the equations x=(t+1)^2 and y=(t+1)^-2 show that the path of the particle is a rectangle hyperbola and determine the velocity and acceleration when t=0
1
Expert's answer
2020-12-22T05:06:32-0500

1) y=1/(t+1)2=1/xy=1/(t+1)^2=1/x . So, y=1/xy=1/x - hyperbola equation in the Cartesian plane.


2) vx=2(t+1)v_x=2(t+1). When t=0t=0 vx=2v_x=2


vy=2(t+1)3v_y=-2(t+1)^{-3}. When t=0t=0 vy=2v_y=-2


v=vx2+vy2=22+(2)2=2.83v=\sqrt{v_x^2+v_y^2}=\sqrt{2^2+(-2)^2}=2.83


3) ax=2a_x=2 and ay=6(t+1)4a_y=6(t+1)^4


When t=0t=0 ax=2a_x=2 and ay=6a_y=6.


a=ax2+ay2=22+62=6.32a=\sqrt{a_x^2+a_y^2}=\sqrt{2^2+6^2}=6.32








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