## Thomas' Calculus 13th Edition

$y= x^2-2x$; $v(1)=i+2j$ and $a(1)=2j$
We know that the velocity and acceleration is given by: velocity is : $v(t)=r'(t)$ and acceleration is: $a(t)=v'(t)$ In the given problem, $x=t+1$ and, $y=t^2-1=(t-1)(t+1)$ Re-write as: $y=(x-2)x \implies x^2-2x$ Now, $v(t)=i+2tj \implies v(1)=i+2j$ we know that acceleration $a(t)=2j \implies a(1)=2j$ Thus, we have $y= x^2-2x$; $v(1)=i+2j$ and $a(1)=2j$