Find the direction and magnitude of the following vectors.
A⃗ =( 22 m )x^+( -15 m )y^
Express your answer in degrees using three significant figures.
θA =
Part B
Express your answer in meters using two significant figures.
|A⃗ | = m
Part C
B⃗ =( 2.5 m )x^+( 16 m )y^
Express your answer in meters using two significant figures.
θB =
Part D
Express your answer in meters using two significant figures.
|B⃗ | = m
Part E
A⃗ +B⃗
Express your answer in degrees using three significant figures.
θA⃗ +B⃗ =
Part F
Express your answer in meters using two significant figures.
|A⃗ +B⃗ | = m
Part A:
To find the direction (θA) and magnitude (|A⃗|) of vector A⃗ =( 22 m )x^+( -15 m )y^, we can use trigonometry.
The direction (θA) can be found using the inverse tangent function:
θA = tan^(-1)(-15 m / 22 m) ≈ -34.9°
The magnitude (|A⃗|) can be found using the Pythagorean theorem:
|A⃗| = √((22 m)^2 + (-15 m)^2) ≈ 26.2 m
Therefore, θA = -34.9° and |A⃗| = 26.2 m.
Part B:
Expressing the magnitude |A⃗| using two significant figures, we get:
|A⃗| ≈ 26 m
Part C:
To find the direction (θB) and magnitude (|B⃗|) of vector B⃗ =( 2.5 m )x^+( 16 m )y^, we can use the same approach as in Part A.
θB = tan^(-1)(16 m / 2.5 m) ≈ 81.9°
|B⃗| = √((2.5 m)^2 + (16 m)^2) ≈ 16.4 m
Therefore, θB = 81.9° and |B⃗| = 16.4 m.
Part D:
Expressing the magnitude |B⃗| using two significant figures, we get:
|B⃗| ≈ 16 m
Part E:
To find the direction (θA⃗ +B⃗), we can add the x-components and y-components of A⃗ and B⃗ separately using vector addition.
(22 m)x^ + (2.5 m)x^ = 24.5 m)x^
+ (-15 m)y^ + (16 m)y^ = 1 m)y^
So, A⃗ +B⃗ = (24.5 m)x^ + (1 m)y^
Using the same approach as in Part A, we can find the direction (θA⃗ +B⃗):
θA⃗ +B⃗ = tan^(-1)((1 m) / (24.5 m)) ≈ 2.3°
Part F:
To find the magnitude |A⃗ +B⃗|, we can use the Pythagorean theorem:
|A⃗ +B⃗| = √((24.5 m)^2 + (1 m)^2) ≈ 24.5 m
Therefore, |A⃗ +B⃗| ≈ 24 m.
To find the direction and magnitude of a vector A⃗ =(22 m)x^+(-15 m)y^, we can use the following formulas:
θA = tan^(-1)(Ay/Ax)
|A⃗ | = √(Ax^2 + Ay^2)
Step 1: Finding θA
Let's substitute the given values:
θA = tan^(-1)(-15 m/22 m)
θA ≈ -34.804 degrees (rounded to three significant figures)
Step 2: Finding |A⃗ |
Substituting the given values:
|A⃗ | = √((22 m)^2 + (-15 m)^2)
|A⃗ | ≈ 26.907 m (rounded to two significant figures)
Now let's apply the same steps for vector B⃗ =(2.5 m)x^+(16 m)y^.
θB = tan^(-1)(By/Bx)
|B⃗ | = √(Bx^2 + By^2)
Step 1: Finding θB
Substituting the given values:
θB = tan^(-1)(16 m/2.5 m)
θB ≈ 81.869 degrees (rounded to three significant figures)
Step 2: Finding |B⃗ |
Substituting the given values:
|B⃗ | = √((2.5 m)^2 + (16 m)^2)
|B⃗ | ≈ 16.281 m (rounded to two significant figures)
For part E, let's add vectors A⃗ and B⃗:
A⃗ + B⃗ = (22 m)x^ + (-15 m)y^ + (2.5 m)x^ + (16 m)y^
= (22 m + 2.5 m)x^ + (-15 m + 16 m)y^
= (24.5 m)x^ + (1 m)y^
To find the direction of A⃗ + B⃗ , we can use the formula:
θA⃗ + B⃗ = tan^(-1)((ΣBy)/(ΣBx))
Substituting the values of ΣBx and ΣBy:
θA⃗ + B⃗ = tan^(-1)((1 m)/(24.5 m))
θA⃗ + B⃗ ≈ 2.313 degrees (rounded to three significant figures)
For part F, let's find |A⃗ + B⃗ |:
|A⃗ + B⃗ | = √((24.5 m)^2 + (1 m)^2)
|A⃗ + B⃗ | ≈ 24.514 m (rounded to two significant figures)
To find the direction and magnitude of a vector, we can use trigonometry. The direction of a vector is given by the angle it makes with the positive x-axis, and the magnitude of a vector is the length of the vector.
Part A:
To find the direction (θA), we use the formula θA = tan^(-1)(Ay/Ax), where Ay is the y-component of the vector and Ax is the x-component of the vector.
Here, Ay = -15 m and Ax = 22 m.
θA = tan^(-1)(-15/22)
To find this value using a calculator, follow these steps:
1. Press the "tan^(-1)" or "inverse tan" button on your calculator.
2. Divide -15 by 22 and enter the result.
3. Press the "=" button to get the angle in radians.
4. To convert radians to degrees, multiply the result by 180/π (approximately 57.2958).
θA ≈ -35.4 degrees (rounded to three significant figures).
To find the magnitude (|A|) of the vector A, we use the formula |A| = sqrt(Ax^2 + Ay^2).
Here, Ax = 22 m and Ay = -15 m.
|A| ≈ sqrt((22^2) + (-15^2))
|A| ≈ sqrt(484 + 225)
|A| ≈ sqrt(709)
|A| ≈ 26.65 m (rounded to two significant figures).
Part C:
To find the direction (θB), we use the same formula as in Part A: θB = tan^(-1)(By/Bx), where By is the y-component of the vector and Bx is the x-component of the vector.
Here, By = 16 m and Bx = 2.5 m.
θB = tan^(-1)(16/2.5)
To find this value using a calculator, follow the same steps as in Part A.
θB ≈ 81.9 degrees (rounded to three significant figures).
To find the magnitude (|B|) of the vector B, we use the same formula as in Part A: |B| = sqrt(Bx^2 + By^2).
Here, Bx = 2.5 m and By = 16 m.
|B| ≈ sqrt((2.5^2) + (16^2))
|B| ≈ sqrt(6.25 + 256)
|B| ≈ sqrt(262.25)
|B| ≈ 16.18 m (rounded to two significant figures).
Part E:
To find the resulting vector when adding two vectors together, we sum their x-components and y-components separately.
Here, A⃗ =( 22 m )x^+( -15 m )y^ and B⃗ =( 2.5 m )x^+( 16 m )y^.
A⃗ + B⃗ = (22 m + 2.5 m) x^ + (-15 m + 16 m) y^
A⃗ + B⃗ = (24.5 m) x^ + (1 m) y^
To find the direction (θA⃗ + B⃗ ), we use the same formula as in Part A: θA⃗ + B⃗ = tan^(-1)(Cy/Cx), where Cy is the y-component of the resulting vector and Cx is the x-component of the resulting vector.
Here, Cx = 24.5 m and Cy = 1 m.
θA⃗ + B⃗ = tan^(-1)(1/24.5)
To find this value using a calculator, follow the same steps as in Part A.
θA⃗ + B⃗ ≈ 2.31 degrees (rounded to three significant figures).
To find the magnitude (|A⃗ + B⃗ |) of the resulting vector, we use the same formula as in Part A: |A⃗ + B⃗ | = sqrt(Cx^2 + Cy^2).
Here, Cx = 24.5 m and Cy = 1 m.
|A⃗ + B⃗ | ≈ sqrt((24.5^2) + (1^2))
|A⃗ + B⃗ | ≈ sqrt(600.25 + 1)
|A⃗ + B⃗ | ≈ sqrt(601.25)
|A⃗ + B⃗ | ≈ 24.52 m (rounded to two significant figures).
I am not going to just do these for you
in general
magnitude = sqrt (xcomponent^2+ycomponent^2)
tan angle from x axis = ycomponent/xcomponent)
to do A + B
say
C = (Ax+Bx)i + (Ay+By)j
where i and j are unit vectors in x and y directions