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. \\(\left|\vec{r}\
ight| = ?\\) miles round your answer to the nearest hundredth.

Explanation:

Step1: Identify the triangle sides and angle

The cyclist's 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, first vector is north (let's say along positive y - axis), second vector is \(50^\circ\) east of north, so the angle between the two vectors is \(50^\circ\)? Wait, no. Let's model the vectors. Let the first vector \(\vec{v_1}\) be \(2\) miles north (so in component form, \(\vec{v_1}=(0, 2)\)). The second vector \(\vec{v_2}\) has magnitude \(5\) miles and direction \(50^\circ\) east of north. So its components are \(v_{2x}=5\sin(50^\circ)\) and \(v_{2y}=5\cos(50^\circ)\). The resultant vector \(\vec{R}=\vec{v_1}+\vec{v_2}\), so \(R_x = 5\sin(50^\circ)\) and \(R_y=2 + 5\cos(50^\circ)\). Then the magnitude of \(\vec{R}\) is \(\sqrt{R_x^{2}+R_y^{2}}\). Alternatively, we can use the Law of Cosines. The angle between the two vectors: the first vector is north, the second is \(50^\circ\) east of north, so the angle between them is \(50^\circ\)? Wait, no. Wait, if we consider the two displacement vectors, the first is length \(a = 2\), the second is length \(b = 5\), and the angle between them \(\theta=180^\circ - 50^\circ= 130^\circ\)? No, that's wrong. Wait, when you go north, then turn \(50^\circ\) east of north, the angle between the initial north direction and the second direction is \(50^\circ\), so the angle between the two vectors (when placed tail - to - tail) is \(180^\circ - 50^\circ= 130^\circ\)? Wait, no. Let's think of the first vector as from point A to point B (2 miles north), then from point B to point C (5 miles at \(50^\circ\) east of north). So the angle at point B between BA and BC: BA is south (opposite of north), and BC is \(50^\circ\) east of north. So the angle between BA (south) and BC (50° east of north) is \(180^\circ - 50^\circ = 130^\circ\). So in triangle ABC, sides AB = 2, BC = 5, and angle at B is \(130^\circ\), and we want to find AC (the magnitude of the resultant vector). So by Law of Cosines, \(AC^{2}=AB^{2}+BC^{2}-2\cdot AB\cdot BC\cdot\cos(180^\circ - 130^\circ)\)? No, wait, Law of Cosines is \(c^{2}=a^{2}+b^{2}-2ab\cos(C)\), where \(C\) is the angle opposite side \(c\). Wait, in our case, the two vectors are \(\vec{AB}\) (2 miles north) and \(\vec{BC}\) (5 miles at \(50^\circ\) east of north). To find the resultant \(\vec{AC}=\vec{AB}+\vec{BC}\), the angle between \(\vec{AB}\) and \(\vec{BC}\) is actually \(180^\circ - 50^\circ= 130^\circ\)? Wait, no. Let's use components.

First vector: \(\vec{v_1}=(0,2)\) (since it's 2 miles north, x - component is 0, y - component is 2).

Second vector: \(\vec{v_2}\): the angle with the y - axis (north) is \(50^\circ\) east, so the x - component is \(5\sin(50^\circ)\) and the y - component is \(5\cos(50^\circ)\). So \(\vec{v_2}=(5\sin(50^\circ),5\cos(50^\circ))\).

Resultant vector \(\vec{R}=\vec{v_1}+\vec{v_2}=(5\sin(50^\circ),2 + 5\cos(50^\circ))\).

Now, calculate the magnitude:

\(R_x = 5\sin(50^\circ)\approx5\times0.7660 = 3.83\)

\(R_y=2 + 5\cos(50^\circ)\approx2+5\times0.6428 = 2 + 3.214=5.214\)

Then \(|\vec{R}|=\sqrt{R_x^{2}+R_y^{2}}=\sqrt{(3.83)^{2}+(5.214)^{2}}=\sqrt{14.6689 + 27.2058}=\sqrt{41.8747}\approx6.47\)

Wait, but let's use Law of Cosines correctly. Let's consider the two vectors: the first has length \(a = 2\), the second has length \(b = 5\), and the angle between them \(\theta = 180^\circ- 50^\circ= 130^\circ\)? No, that's incorrect. Wait, when you have two vectors, one is north (length 2), the other is \(50^\circ\) east of north (length 5). The angle between them (when placed tail - to - tail) is \(50^\circ\), not \(130^\circ\). Wait, I made a mistake earlier. Let's place the two vectors tail - to - tail at the starting point. The first vector is along the positive y - axis (north), the second vector is at an angle of \(50^\circ\) from the positive y - axis towards the positive x - axis (east). So the angle between the two vectors \(\theta = 50^\circ\). Then by the Law of Cosines, the magnitude of the resultant \(R\) is given by \(R^{2}=a^{2}+b^{2}+2ab\cos\theta\)? Wait, no. Wait, the Law of Cosines for vector addition: if \(\vec{R}=\vec{A}+\vec{B}\), then \(|\vec{R}|^{2}=|\vec{A}|^{2}+|\vec{B}|^{2}+2|\vec{A}||\vec{B}|\cos\theta\), where \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\) when placed tail - to - tail. In this case, \(\vec{A}\) is 2 miles north, \(\vec{B}\) is 5 miles at \(50^\circ\) east of north. So the angle between \(\vec{A}\) and \(\vec{B}\) is \(50^\circ\), so \(R^{2}=2^{2}+5^{2}+2\times2\times5\times\cos(50^\circ)\). Let's calculate that. \(\cos(50^\circ)\approx0.6428\). So \(R^{2}=4 + 25+20\times0.6428=29 + 12.856 = 41.856\). Then \(R=\sqrt{41.856}\approx6.47\). Yes, that matches the component method.

Step2: Calculate using Law of Cosines

We have two sides of a triangle (the two displacement vectors) with lengths \(a = 2\) and \(b = 5\), and the included angle \(\theta=50^\circ\) (wait, no, when adding vectors \(\vec{R}=\vec{A}+\vec{B}\), the formula is \(|\vec{R}|^{2}=|\vec{A}|^{2}+|\vec{B}|^{2}+2|\vec{A}||\vec{B}|\cos\theta\), where \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\) (tail - to - tail). In this case, \(\vec{A}\) is 2 miles north, \(\vec{B}\) is 5 miles at \(50^\circ\) east of north, so the angle between them is \(50^\circ\). So:

\(|\vec{R}|^{2}=2^{2}+5^{2}+2\times2\times5\times\cos(50^\circ)\)

\(=4 + 25+20\times\cos(50^\circ)\)

\(\cos(50^\circ)\approx0.6427876097\)

\(20\times\cos(50^\circ)\approx20\times0.6427876097 = 12.85575219\)

\(|\vec{R}|^{2}=29 + 12.85575219=41.85575219\)

\(|\vec{R}|=\sqrt{41.85575219}\approx6.47\)

Answer:

\(6.47\)