Question

a cyclist goes 2 miles north, then turns and goes at an angle of 50° east of north for 5 miles. what is the magnitude of the cyclists resultant vector? draw a vector diagram. |\overrightarrow{r}| = ? miles round your answer to the nearest hundredth.

Explanation:

Step1: Identify the triangle sides and angle

The first displacement is 2 miles north, the second is 5 miles at \(50^\circ\) east of north. The angle between the two vectors is \(180^\circ - 50^\circ= 130^\circ\)? Wait, no. Wait, when moving north, then \(50^\circ\) east of north, the angle between the two vectors (the initial north vector and the second vector) is actually \(180^\circ - 130^\circ\)? Wait, no. Let's think again. The first vector is along the north direction (let's say along the positive y - axis). The second vector is at an angle of \(50^\circ\) east of north, so the angle between the two vectors (the angle between the first vector \(\vec{v_1}\) with magnitude \(v_1 = 2\) and the second vector \(\vec{v_2}\) with magnitude \(v_2=5\)) is \(180^\circ-(90^\circ + 50^\circ)\)? No, wait, no. If the first vector is north (let's define north as the positive y - direction), and the second vector is \(50^\circ\) east of north, then the angle between the two vectors is \(180^\circ - 130^\circ\)? Wait, no. Let's use the law of cosines. The law of cosines for the magnitude of the resultant vector \(\vec{R}\) when we have two vectors \(\vec{A}\) and \(\vec{B}\) with magnitudes \(a\) and \(b\) and the angle \(\theta\) between them is \(|\vec{R}|=\sqrt{a^{2}+b^{2}-2ab\cos(180^{\circ}-\theta)}\)? Wait, no. Wait, the angle between the two vectors: the first vector is 2 miles north, the second is 5 miles at \(50^\circ\) east of north. So the angle between the two vectors (the angle inside the triangle formed by the two vectors and the resultant) is \(180^\circ - 50^\circ=130^\circ\)? Wait, no. Let's draw a mental picture. The first leg: 2 miles north. The second leg: 5 miles at \(50^\circ\) east of north. So the angle between the two vectors (the angle between the 2 - mile vector and the 5 - mile vector) is \(180^\circ - 130^\circ\)? No, actually, when you have the first vector going north, and the second vector going \(50^\circ\) east of north, the angle between the two vectors (the angle between the two sides of length 2 and 5) is \(180^\circ-(90^\circ - 50^\circ)\)? I think I made a mistake. Let's use the law of cosines correctly. Let's denote the first vector as \(A = 2\) (magnitude), the second as \(B = 5\) (magnitude). The angle between them: when moving north, then \(50^\circ\) east of north, the angle between the two vectors is \(180^\circ - 50^\circ = 130^\circ\)? Wait, no. If the first vector is along the positive y - axis (north), and the second vector is at an angle of \(50^\circ\) from the north towards the east, then the angle between the two vectors (the angle between the first vector and the second vector) is \(180^\circ - 130^\circ\)? No, the angle between the two vectors is actually \(180^\circ - 130^\circ\)? Wait, no. Let's use the law of cosines. The formula for the magnitude of the resultant of two vectors \(\vec{A}\) and \(\vec{B}\) with magnitudes \(a\) and \(b\) and the angle \(\theta\) between them is \(|\vec{R}|=\sqrt{a^{2}+b^{2}+2ab\cos\theta}\), where \(\theta\) is the angle between the two vectors when they are placed tail - to - tail. If we place the two vectors tail - to - tail, the first vector is north, the second vector is \(50^\circ\) east of north, so the angle between them is \(180^\circ - 130^\circ\)? Wait, no. Let's consider the angle between the two vectors. The first vector is along the north direction (let's say the positive y - axis). The second vector is at an angle of \(50^\circ\) from the north towards the east. So the angle between the two vectors (the angle between the first vector and the second vector when they are placed tail - to - tail) is \(180^\circ-(90^\circ + 50^\circ)\)? No, that's not right. Wait, if the first vector is north (y - axis), and the second vector is \(50^\circ\) east of north, then the angle between the two vectors is \(180^\circ - 130^\circ\)? Wait, I think I messed up. Let's use the law of cosines correctly. Let's define the two vectors:

Vector 1: \(\vec{v_1}\) with magnitude \(v_1 = 2\), direction north (let's say \(\theta_1 = 90^\circ\) from the positive x - axis).

Vector 2: \(\vec{v_2}\) with magnitude \(v_2 = 5\), direction \(90^\circ- 50^\circ=40^\circ\) from the positive x - axis (since \(50^\circ\) east of north: north is \(90^\circ\) from x - axis, so \(90^\circ - 50^\circ = 40^\circ\) from x - axis).

The angle between the two vectors is \(90^\circ - 40^\circ=50^\circ\)? No, that's not. Wait, the angle between \(\vec{v_1}\) (at \(90^\circ\) from x - axis) and \(\vec{v_2}\) (at \(40^\circ\) from x - axis) is \(90^\circ - 40^\circ = 50^\circ\)? No, the angle between two vectors with angles \(\theta_1\) and \(\theta_2\) from the x - axis is \(|\theta_1-\theta_2|\). So \(\theta_1 = 90^\circ\), \(\theta_2=90^\circ - 50^\circ = 40^\circ\), so the angle between them is \(|90^\circ - 40^\circ| = 50^\circ\)? Wait, that can't be. If the first vector is north (up the y - axis) and the second is \(50^\circ\) east of north, then the angle between them is \(50^\circ\). Wait, yes! Because from the north direction (the first vector), turning \(50^\circ\) towards the east (the second vector), so the angle between the two vectors is \(50^\circ\). Wait, no, that's the angle of the second vector from the first vector. So when we want to find the resultant, we can use the law of cosines where the two sides are \(a = 2\), \(b = 5\), and the included angle \(\theta=180^\circ - 50^\circ = 130^\circ\)? Wait, I'm getting confused. Let's use the law of cosines formula. The law of cosines states that for a triangle with sides \(a\), \(b\), and \(c\), and the angle \(\gamma\) opposite side \(c\), \(c^{2}=a^{2}+b^{2}-2ab\cos\gamma\). In the case of vector addition, if we have two vectors \(\vec{A}\) and \(\vec{B}\), the magnitude of the resultant \(\vec{R}=\vec{A}+\vec{B}\) is given by \(|\vec{R}|=\sqrt{|\vec{A}|^{2}+|\vec{B}|^{2}+2|\vec{A}||\vec{B}|\cos\theta}\), where \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\) when they are placed tail - to - tail.

Wait, let's take a step back. Let's model the two displacements:

- Displacement 1: \(d_1 = 2\) miles, direction: north (let's represent this as a vector along the positive y - axis: \(\vec{d_1}=(0,2)\))

- Displacement 2: \(d_2 = 5\) miles, direction: \(50^\circ\) east of north. To find the components of this vector:

The x - component (east - west) of \(\vec{d_2}\) is \(d_{2x}=5\sin(50^\circ)\) (since east is the positive x - direction, and the angle with the north (y - axis) is \(50^\circ\), so \(\sin(50^\circ)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{d_{2x}}{5}\))

The y - component (north - south) of \(\vec{d_2}\) is \(d_{2y}=5\cos(50^\circ)\) (since \(\cos(50^\circ)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{d_{2y}}{5}\))

Now, the resultant vector \(\vec{R}\) has components:

\(R_x=d_{2x}=5\sin(50^\circ)\)

\(R_y=d_{1y}+d_{2y}=2 + 5\cos(50^\circ)\)

Then, the magnitude of the resultant vector \(|\vec{R}|=\sqrt{R_x^{2}+R_y^{2}}\)

Let's calculate the components:

First, calculate \(R_x\):

\(\sin(50^\circ)\approx0.7660\), so \(R_x = 5\times0.7660 = 3.83\)

Then, calculate \(R_y\):

\(\cos(50^\circ)\approx0.6428\), so \(R_y=2 + 5\times0.6428=2 + 3.214 = 5.214\)

Now, calculate the magnitude of \(\vec{R}\):

\(|\vec{R}|=\sqrt{(3.83)^{2}+(5.214)^{2}}\)

\((3.83)^{2}=3.83\times3.83 = 14.6689\)

\((5.214)^{2}=5.214\times5.214\approx27.1858\)

Sum of squares: \(14.6689 + 27.1858=41.8547\)

Take the square root: \(\sqrt{41.8547}\approx6.47\)

Wait, but let's use the law of cosines correctly. The angle between the two vectors: the first vector is 2 miles north, the second is 5 miles at \(50^\circ\) east of north. So the angle between the two vectors (the angle inside the triangle formed by the two vectors and the resultant) is \(180^\circ - 50^\circ = 130^\circ\)? Wait, no. Wait, if we consider the two vectors as two sides of a triangle, and the resultant as the third side, the angle between the two vectors (the angle between the 2 - mile side and the 5 - mile side) is \(180^\circ - 50^\circ=130^\circ\). Let's apply the law of cosines:

\(|\vec{R}|=\sqrt{2^{2}+5^{2}-2\times2\times5\times\cos(130^\circ)}\)

First, calculate \(\cos(130^\circ)\). Since \(130^\circ\) is in the second quadrant, \(\cos(130^\circ)=-\cos(50^\circ)\approx - 0.6428\)

So,

\(|\vec{R}|=\sqrt{4 + 25-20\times(- 0.6428)}\)

\(=\sqrt{29 + 12.856}\)

\(=\sqrt{41.856}\approx6.47\)

Yes, that matches the component method. So the magnitude of the resultant vector is approximately 6.47 miles.

Step1: Recall the Law of Cosines

The law of cosines for the magnitude of the resultant vector \(|\vec{R}|\) of two vectors \(\vec{A}\) (magnitude \(a = 2\)) and \(\vec{B}\) (magnitude \(b = 5\)) with the angle \(\theta\) between them is given by:
\[|\vec{R}|=\sqrt{a^{2}+b^{2}-2ab\cos(180^{\circ}-\alpha)}\]
Wait, no. Wait, the angle between the two vectors: when the first vector is north and the second is \(50^{\circ}\) east of north, the angle between them (the angle inside the triangle) is \(\theta=180^{\circ}- 50^{\circ}=130^{\circ}\) (because the angle between the two vectors when placed tail - to - tail, and we want the angle for the law of cosines in the triangle formed by the two vectors and the resultant).

Step2: Substitute the values into the Law of Cosines formula

We have \(a = 2\), \(b = 5\), and \(\theta = 130^{\circ}\). The law of cosines formula is:
\[|\vec{R}|=\sqrt{a^{2}+b^{2}-2ab\cos\theta}\]
Substitute \(a = 2\), \(b = 5\), and \(\cos(130^{\circ})\approx - 0.6428\) into the formula:
\[|\vec{R}|=\sqrt{2^{2}+5^{2}-2\times2\times5\times\cos(130^{\circ})}\]
\[=\sqrt{4 + 25-20\times(- 0.6428)}\]

Step3: Calculate the expression inside the square root

First, calculate the products and sums:
\[4 + 25=29\]
\[20\times(-0.6428)=- 12.856\]
\[29-20\times(-0.6428)=29 + 12.856 = 41.856\]

Step4: Take the square root

\[|\vec{R}|=\sqrt{41.856}\approx6.47\]

Answer:

\(6.47\)