Question

find the magnitude of the vector difference \\(\vec{a} - \vec{b}\\). express your answer in meters.

Explanation:

Step1: Determine components of \(\vec{A}\) and \(\vec{B}\)

• For \(\vec{A}\) (magnitude \(A = 8.00\) m, direction: negative x - axis, so \(\theta_A=180^\circ\)):
• \(A_x = A\cos\theta_A=8.00\cos180^\circ=- 8.00\) m
• \(A_y = A\sin\theta_A=8.00\sin180^\circ = 0\) m
• For \(\vec{B}\) (magnitude \(B = 15.0\) m, direction: \(\theta_B = 30.0^\circ\) below positive x - axis, so \(\theta_B=- 30.0^\circ\) or \(330^\circ\)):
• \(B_x = B\cos\theta_B=15.0\cos(-30.0^\circ)=15.0\times\frac{\sqrt{3}}{2}\approx12.99\) m
• \(B_y = B\sin\theta_B=15.0\sin(-30.0^\circ)=15.0\times(-\frac{1}{2})=-7.50\) m

Step2: Find components of \(\vec{A}-\vec{B}\)

• The x - component of \(\vec{A}-\vec{B}\) is \( (A - B)_x=A_x - B_x\)
• \( (A - B)_x=-8.00-12.99=-20.99\) m
• The y - component of \(\vec{A}-\vec{B}\) is \( (A - B)_y=A_y - B_y\)
• \( (A - B)_y=0-(-7.50) = 7.50\) m

Step3: Calculate the magnitude of \(\vec{A}-\vec{B}\)

The magnitude of a vector \(\vec{V}\) with components \(V_x\) and \(V_y\) is given by \(|\vec{V}|=\sqrt{V_x^2 + V_y^2}\)
For \(\vec{A}-\vec{B}\), \(|\vec{A}-\vec{B}|=\sqrt{(-20.99)^2+(7.50)^2}\)
First, calculate \((-20.99)^2\approx440.58\) and \((7.50)^2 = 56.25\)
Then, \(440.58 + 56.25=496.83\)
\(|\vec{A}-\vec{B}|=\sqrt{496.83}\approx22.3\) m

Answer:

\(22.3\)