Question

use your answers to find the magnitude of $-vec{a} - vec{b}$. express your answer in meters. $| -vec{a} - vec{b} | = \boxed{?}$ m (diagram shows vectors: $vec{a}$ (8.00 m), $vec{b}$ (15.0 m), $vec{c}$ (12.0 m), $vec{d}$ (10.0 m) with angles 25.0°, 53.0°, 30.0° relative to axes)

Explanation:

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

\(\vec{A}\) is along the negative x - axis, so \(A_x=- 8.00\space m\), \(A_y = 0\space m\).
For \(\vec{B}\), the angle with the positive x - axis is \(- 30.0^{\circ}\) (or \(330^{\circ}\)). Using the component formulas \(B_x=B\cos\theta\) and \(B_y = B\sin\theta\), where \(B = 15.0\space m\) and \(\theta=-30.0^{\circ}\).
\(B_x=15.0\cos(- 30.0^{\circ})=15.0\times\frac{\sqrt{3}}{2}\approx12.99\space m\)
\(B_y=15.0\sin(-30.0^{\circ})=15.0\times(-\frac{1}{2})=- 7.50\space m\)

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

The x - component of \(-\vec{A}-\vec{B}\) is \((-A_x)-B_x\). Since \(A_x=-8.00\), \(-A_x = 8.00\). So \((- \vec{A}-\vec{B})_x=8.00 - 12.99=-4.99\space m\)
The y - component of \(-\vec{A}-\vec{B}\) is \((-A_y)-B_y\). Since \(A_y = 0\), \((-\vec{A}-\vec{B})_y=0-(-7.50) = 7.50\space m\)

Step3: Calculate the magnitude

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 \(-\vec{A}-\vec{B}\), \(|\vec{-A}-\vec{B}|=\sqrt{(-4.99)^{2}+(7.50)^{2}}\)
\((-4.99)^{2}\approx24.90\), \((7.50)^{2}=56.25\)
\(|\vec{-A}-\vec{B}|=\sqrt{24.90 + 56.25}=\sqrt{81.15}\approx9.01\space m\) (more accurately, let's recalculate the components with more precision)

Let's re - calculate the components:
\(\vec{A}\): \(A_x=-8.00\space m\), \(A_y = 0\)
\(\vec{B}\): \(\theta=-30^{\circ}\), \(B = 15.0\space m\)
\(B_x=15.0\times\cos(-30^{\circ})=15.0\times\frac{\sqrt{3}}{2}=15.0\times0.8660 = 12.99\space m\)
\(B_y=15.0\times\sin(-30^{\circ})=15.0\times(-0.5)=-7.50\space m\)

\(-\vec{A}\): \(A'_x = 8.00\space m\), \(A'_y=0\space m\)
\(-\vec{A}-\vec{B}\): \(x\) - component: \(8.00-12.99=-4.99\space m\)
\(y\) - component: \(0 - (-7.50)=7.50\space m\)

Magnitude: \(\sqrt{(-4.99)^{2}+(7.50)^{2}}=\sqrt{24.9001 + 56.25}=\sqrt{81.1501}\approx9.01\space m\) (if we use more precise calculation, let's consider \(\vec{A}\) is along negative x, so \(-\vec{A}\) is along positive x with magnitude \(8.00\space m\), \(\vec{B}\) has components \(B_x = 15\cos(-30^{\circ})\), \(B_y=15\sin(-30^{\circ})\). Then \(-\vec{A}-\vec{B}=(-\vec{A})+(-\vec{B})\). The vector \(-\vec{B}\) has components \(-B_x=-15\cos(-30^{\circ})\), \(-B_y=-15\sin(-30^{\circ})\)

So \((-\vec{A}-\vec{B})_x=8.00-15\cos(-30^{\circ})\), \((-\vec{A}-\vec{B})_y=0 - 15\sin(-30^{\circ})\)

\(\cos(-30^{\circ})=\frac{\sqrt{3}}{2}\approx0.8660\), \(\sin(-30^{\circ})=- 0.5\)

\((-\vec{A}-\vec{B})_x=8.00-15\times0.8660=8.00 - 12.99 = - 4.99\)

\((-\vec{A}-\vec{B})_y=0-15\times(-0.5)=7.5\)

Then magnitude \(=\sqrt{(-4.99)^2 + 7.5^2}=\sqrt{24.9001+56.25}=\sqrt{81.1501}\approx9.01\space m\). If we calculate \(\sqrt{(-5)^2+7.5^2}=\sqrt{25 + 56.25}=\sqrt{81.25}\approx9.01\space m\) (since \(-4.99\approx - 5\))

Answer:

\(\approx9.01\space m\) (or more precisely, after exact calculation, it is approximately \(9.0\space m\) or \(9.01\space m\))