An expedition walks 4 kilometers from camp on a bearing of 30 degrees, then turns and walks 10 km on a bearing of 160 degrees. Find the magnitude of the new location relative to the beginning location using vectors.
-> if you can, can you possibly show how the triangle would look like along with how to solve it using vectors/ vector formula.
-> here's what I have so far: v1=4(cos60, sin60), v2=10(cos70, -sin70), v1+v2=vr, ||vr||=sqrt of vectors^2 from vr
Sure, let's break it down step by step using vectors.
First, let's represent the initial displacement from the camp using vector v1. We know that it is 4 kilometers on a bearing of 30 degrees. To represent this as a vector, we can use the formula v1 = 4(cos30°, sin30°).
Next, let's represent the second displacement from the end of v1 using vector v2. We know that it is 10 kilometers on a bearing of 160 degrees. To represent this as a vector, we can use the formula v2 = 10(cos160°, sin160°).
Now, let's sum up both vectors v1 and v2 to find the resultant vector vr. We can add them component-wise, by adding the x-components and y-components separately. So, vr = (v1x + v2x, v1y + v2y).
Let's calculate the components of vr:
v1x = 4 * cos30° ≈ 3.464
v1y = 4 * sin30° ≈ 2
v2x = 10 * cos160° ≈ -9.659
v2y = 10 * sin160° ≈ -2.588
Now, let's add the components to find vr:
vr ≈ (3.464 + (-9.659), 2 + (-2.588)) ≈ (-6.196, -0.588)
Finally, we can find the magnitude of the resultant vector vr using the formula ||vr|| = √(vr_x^2 + vr_y^2). Let's calculate it:
||vr|| = √((-6.196)^2 + (-0.588)^2)
≈ √(38.428 + 0.345)
≈ √38.773
≈ 6.227 kilometers (rounded to three decimal places)
So, the magnitude of the new location relative to the beginning location is approximately 6.227 kilometers.
To solve this problem using vectors, let's break down the information given step by step:
Step 1: Calculate the displacement vectors.
- The first leg of the journey is 4 kilometers on a bearing of 30 degrees. To represent this as a displacement vector, we can use vector v1 = 4(cos30, sin30).
- The second leg of the journey is 10 kilometers on a bearing of 160 degrees. To represent this as a displacement vector, we can use vector v2 = 10(cos160, sin160).
Step 2: Find the resultant vector (vr) by adding v1 and v2.
- vr = v1 + v2
- vr = 4(cos30, sin30) + 10(cos160, sin160)
- vr = (4cos30 + 10cos160, 4sin30 + 10sin160)
Step 3: Calculate the magnitude of the resultant vector.
- The magnitude (also known as the length or magnitude) of vr can be found using the formula:
||vr|| = sqrt((4cos30 + 10cos160)^2 + (4sin30 + 10sin160)^2)
Step 4: Simplify the equation and calculate the magnitude of vr.
- Evaluate the trigonometric functions and calculate the magnitude of vr using the formula above.
By following these steps, you'll be able to calculate the magnitude of the new location relative to the beginning location using vectors.
To solve this problem using vectors, we need to break down the information provided and use vector addition.
Step 1: Visualize the Triangle
Let's draw a triangle to represent the expedition's movements. Start by drawing a point to represent the camp (let's call it point A). From point A, draw a vector representing the first leg of the expedition, which is 4 km at a bearing of 30 degrees. Then, draw another vector from the endpoint of the first leg to represent the second leg of the expedition, which is 10 km at a bearing of 160 degrees. Finally, connect the endpoint of the second leg back to the starting point, forming a triangle.
Step 2: Break Down the Vectors
Now, let's break down the vectors into their x and y components.
The first vector, v1, represents the 4 km at a bearing of 30 degrees. To find the x and y components, we can use trigonometry. Since 30 degrees is an acute angle, we can use the cosine and sine of 30 degrees to find the x and y components, respectively.
v1 = 4(cos(30), sin(30))
= 4(√3/2, 1/2)
= (2√3, 2)
The second vector, v2, represents the 10 km at a bearing of 160 degrees. To find the x and y components, we again use trigonometry. Since 160 degrees is an obtuse angle, we can use the cosine and sine of (180 - 160 = 20) degrees to find the x and y components, respectively.
v2 = 10(cos(20), -sin(20))
= 10(cos(20), -√3/2)
= (10cos(20), -5√3)
Step 3: Add the Vectors
To find the vector representing the expedition's displacement from the starting point to the ending point, we add the x and y components of v1 and v2.
vr = v1 + v2
= (2√3 + 10cos(20), 2 - 5√3)
Step 4: Calculate the Magnitude
To find the magnitude of the displacement vector vr, we can use the formula ||vr|| = √(x^2 + y^2).
||vr|| = √((2√3 + 10cos(20))^2 + (2 - 5√3)^2)
Now, you can use a calculator to evaluate this expression and find the magnitude of the displacement vector.
Disp. = 4km[30o] + 10km[160o].
X = 4*Cos30 + 10*Cos160 = -5.93km.
Y = 4*sin30 + 10*sin160 = 5.42km.
Tan A = Y/X = 5.42/(-5.93) = -0.91403, A = -42.43o N. of W.137.6o CCW from + x-axis.
Disp. = X/Cos A = (-5.93)/Cos137.6 = 8.03 km[137.6]