Question

1. a charged particle is on an x - y plane. the particle starts from the origin, and begins by moving 14.0 centimeters, at an angle of 21.1° below the positive x - axis. the particle then changes direction and travels 11.1 cm at an angle of 38.8° above the positive x - axis. what is the particles displacement vector after these two motions? 21.8 cm, 5.03° above the positive x - axis 17.7 cm, 6.71° below the positive x - axis 17.7 cm, 6.71° above the positive x - axis 21.8 cm, 5.03° below the positive x - axis

Explanation:

Step1: Calculate \(x\) - components of the two vectors

For the first vector \(\vec{A}\) with magnitude \(A = 14.0\space cm\) and angle \(\theta_{A}=- 21.1^{\circ}\) (below the \(x\) - axis), the \(x\) - component is \(A_{x}=A\cos\theta_{A}\).

$$A_{x}=14.0\cos(-21.1^{\circ})=14.0\times\cos(21.1^{\circ})\approx14.0\times0.933 = 13.062\space cm$$

For the second vector \(\vec{B}\) with magnitude \(B = 11.1\space cm\) and angle \(\theta_{B}=38.8^{\circ}\) (above the \(x\) - axis), the \(x\) - component is \(B_{x}=B\cos\theta_{B}\).

$$B_{x}=11.1\cos(38.8^{\circ})\approx11.1\times0.780=8.658\space cm$$

The total \(x\) - component \(R_{x}=A_{x}+B_{x}\)

$$R_{x}=13.062 + 8.658=21.72\space cm$$

Step2: Calculate \(y\) - components of the two vectors

For the first vector \(\vec{A}\), the \(y\) - component is \(A_{y}=A\sin\theta_{A}\).

$$A_{y}=14.0\sin(-21.1^{\circ})=-14.0\times\sin(21.1^{\circ})\approx-14.0\times0.359=-5.026\space cm$$

For the second vector \(\vec{B}\), the \(y\) - component is \(B_{y}=B\sin\theta_{B}\).

$$B_{y}=11.1\sin(38.8^{\circ})\approx11.1\times0.627 = 6.9597\space cm$$

The total \(y\) - component \(R_{y}=A_{y}+B_{y}\)

$$R_{y}=- 5.026+6.9597 = 1.9337\space cm$$

Step3: Calculate the magnitude of the resultant vector \(R\)

Using the Pythagorean theorem \(R=\sqrt{R_{x}^{2}+R_{y}^{2}}\)

$$R=\sqrt{(21.72)^{2}+(1.9337)^{2}}=\sqrt{471.7584 + 3.739}\approx\sqrt{475.497}\approx17.7\space cm$$

Step4: Calculate the angle \(\theta\) of the resultant vector

Using the formula \(\tan\theta=\frac{R_{y}}{R_{x}}\)

$$\tan\theta=\frac{1.9337}{21.72}\approx0.089$$

$$\theta=\arctan(0.089)\approx6.71^{\circ}$$

(above the \(x\) - axis since \(R_{y}>0\))

Answer:

17.7 cm, \(6.71^{\circ}\) above the positive \(x\) - axis (the third option)