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?
$61^{\circ}$
$29^{\circ}$
$72^{\circ}$
$17^{\circ}$
\(\boxed{61^\circ}\)
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 :\(\boxed{61^\circ}\)