A jogger runs 145m. in a direction 20.0 degrees east of north (displacement of vector A) and then 105m. in a direction 35.0 degrees south of east (displacement of vector B). Using components, determine the magnitude and direction of the resultant vector C for these two displacements.
Why making fool give full answer
Thank you. I got now the answer. :)
thanks for the hint. why not the full answer
To determine the magnitude and direction of the resultant vector C, we can break down the given displacements into their respective horizontal and vertical components.
Let's start by calculating the horizontal and vertical components of vector A.
Horizontal component of vector A:
Aᵤx = A * cos(θ)
Aᵤx = 145m * cos(20°)
Aᵤx ≈ 137.66m
Vertical component of vector A:
Aᵥy = A * sin(θ)
Aᵥy = 145m * sin(20°)
Aᵥy ≈ 49.59m
Now, let's calculate the horizontal and vertical components of vector B.
Horizontal component of vector B:
Bᵤx = B * cos(θ)
Bᵤx = 105m * cos(35°)
Bᵤx ≈ 85.89m
Vertical component of vector B:
Bᵥy = B * sin(θ)
Bᵥy = 105m * sin(35°)
Bᵥy ≈ -60.10m (negative because it's south)
To find the resultant vector C, we add the horizontal and vertical components of vectors A and B:
Horizontal component of vector C:
Cᵤx = Aᵤx + Bᵤx
Cᵤx = 137.66m + 85.89m
Cᵤx ≈ 223.55m
Vertical component of vector C:
Cᵥy = Aᵥy + Bᵥy
Cᵥy = 49.59m - 60.10m (since Bᵥy is negative)
Cᵥy ≈ -10.51m
Next, we can calculate the magnitude of vector C using the Pythagorean theorem:
|C| = √(Cᵤx² + Cᵥy²)
|C| = √(223.55m² + (-10.51m)²)
|C| ≈ √(49932.80m²)
|C| ≈ 223.59m (rounded to two decimal places)
Finally, to determine the direction of vector C, we can use the inverse tangent function:
θ = tan^(-1)(Cᵥy / Cᵤx)
θ = tan^(-1)(-10.51m / 223.55m)
θ ≈ -2.68° (rounded to two decimal places)
The magnitude of the resultant vector C is approximately 223.59m, and the direction is approximately 2.68 degrees below the positive x-axis.
Hints:
X-component of resultant
= ∑Di*cos(θi)
and
Y-component of resultant
= ∑Di*sin(θi)
D(1)=145m, θ(1)=90-20=70° (from x-axis)
D(2)=105m, θ(2)=-35° (from x-axis)