3.1 Displacement, Velocity, and Acceleration A car goes from point A to point B. According to its coordinate position in meters as a function of time $t$ in seconds, $x(t)=t^{2}+2.0 t+1.0$ and, $y(t)=3.0 t^{2}+3.0 t-2.0$. What is the angle of the car's position vector after 2.0 s relative to $x$-axis? $61^{\circ}$ $29^{\circ}$ $72^{\circ}$ $17^{\circ}$

Step 1 ：First, we need to find the position vector of the car after 2 seconds. We can do this by plugging in \(t=2\) into the given equations for \(x(t)\) and \(y(t)\).

Step 2 ：\(t = 2\)

Step 3 ：\(x_t = 2^2 + 2(2) + 1 = 9\)

Step 4 ：\(y_t = 3(2)^2 + 3(2) - 2 = 16\)

Step 5 ：Now that we have the position vector \((9, 16)\), we can find the angle between the position vector and the x-axis using the arctangent function. The angle can be calculated as \(\theta = \arctan(\frac{y}{x})\).

Step 6 ：\(\theta = \arctan(\frac{16}{9})\)

Step 7 ：\(\theta \approx 1.0584\)

Step 8 ：\(\theta \approx 60.64^\circ\)

Step 9 ：\(\boxed{61^\circ}\)