Vector A⃗ is 2.80 cm long and is 60.0∘ above the x-axis in the first quadrant….
The following solution is suggested to handle the subject “Vector A⃗ is 2.80 cm long and is 60.0∘ above the x-axis in the first quadrant….“. Let’s keep an eye on the content below!
Question “Vector A⃗ is 2.80 cm long and is 60.0∘ above the x-axis in the first quadrant….”
Vector A⃗ is 2.80 cm long and is 60.0∘ above the x-axis in the
first quadrant. Vector B⃗ is 1.90 cm long and is 60.0∘ below the
x-axis in the fourth quadrant (the figure (Figure 1) Use components
to find the magnitude of B⃗ −A⃗
Use components to find the direction of B⃗ −A⃗ .
Sketch the vector subtraction C⃗ =B⃗
−A⃗ .
Answer
Let’s say that Vector is R, and it makes angle with +x axis. Then its components are:
Rx = Horizontal component = R*cos
Ry = Vertical component =R*sin
Use the following rule:
Vector A = 2.80cm, 60.0 degrees above +x axis. (In the 1st quadrant both vector x and vector y are epositive).
Ax = A*cos = 2.80*cos 60 deg = 1.40 cm
Ay = A*sin = 2.80*sin 60 deg = 2.425 cm
Vector B = 1.90cm, 60.0 degrees below +x axis. (In the 4th quadrant, x is +ve while y is negative.)
Bx = B*cos = 1.90*cos 60 deg = 0.95 cm
By = -B*sin = -1.90*sin 60 deg = -1.645 cm
We now need to figure out
C = B – A = B + (-A).
A = Ax i + Ay j
(-A) = -Ax i – Ay j = -1.40 i – 2.425 j
B = Bx i + By j = 0.95 i – 1.645 j
So,
C = (Bx – Ax) i + (By – Ay) j
C = (0.95 – 1.40) i + (-1.645 – 2.425) j
C = -0.45 i – 4.07 j
The magnitude of vector C will be:
|C| = |B – A|
C
C
Direction of vector C is:
Direction = 180 + arctan (|Cy|/|Cx|)
Direction = 180 + arctan (4.07/0.45) = 180 + 83.7
Direction = 263.7CCW starting at +ve x-axis
Part C.
Please upvote.
Please let me know if there are any questions.
Conclusion
Above is the solution for “Vector A⃗ is 2.80 cm long and is 60.0∘ above the x-axis in the first quadrant….“. We hope that you find a good answer and gain the knowledge about this topic of science.