Answer:
The answer is 135 degrees.
Step-by-step explanation:
As we are given the position. If we take the derivative, we get the velocity vector. If we take the derivative again, we find the acceleration vector of the particle.
[tex]r(t)=(2t)i+(2t-16t^{2})j[/tex]
[tex]V(t)=2i+(2-32t)j[/tex]
[tex]a(t)=-32j[/tex]
At time t=0;
[tex]v(t)=2i+2j[/tex]
[tex]a(t)=-32j[/tex]
As i attach in the picture the angle between the velocity and acceleration vector is [tex](45+90)=135[/tex] degrees
