Question

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

Answer 1

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 $a_y=6$.

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