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

Expert's answer

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

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

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

"v=\\sqrt{v_x^2+v_y^2}=\\sqrt{2^2+(-2)^2}=2.83"

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

When "t=0" "a_x=2" and "a_y=6".

"a=\\sqrt{a_x^2+a_y^2}=\\sqrt{2^2+6^2}=6.32"

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