The displacement vectors A and B shown the figure both have in below magnitudes

Question

The displacement vectors A and B shown the figure both have in below magnitudes of 1.80 m. The direction of vector A is (a) Find A B graphically. 3.6 magnitude Your response differs from the correct answer by more than 10%. Double check your calculations. m direction o counterclockwise from the +x axis (b) Find A B graphically magnitude direction o counterclockwise from the +x axis (c) Find B A graphically magnitude direction o counterclockwise from the +x axis (d) Find A 2B graphically. magnitude direction o counterclockwise from the +x axis 29.2

Explanation / Answer

from the given data,

Ax = A*cos(29.2) = 1.8*cos(29.2) = 1.57 m

Ay = A*sin(19.2) = 1.8*sin(29.2) = 0.89 m

A = 1.57i + 0.89 j

Bx = 0

By = 1.8

A = 0i + 1.8j

A) A + B = (1.57i + 0.89 j) + (0i + 1.8j)

= 1.57i + 2.69 j

|A+B| = sqrt(1.57^2 + 2.69^2) = 3.11 m

direction : theta = tan^-1(2.69/1.57) = 59.73 degrees

B) A - B = (1.57i + 0.89 j) - (0i + 1.8j)

= 1.57i - 0.91 j

|A-B| = sqrt(1.57^2 + 0.91^2) = 1.81 m

direction : theta = tan^-1(-0.91/1.57) = 330 degrees

C)
B - A = -(A - B) = -( (1.57i + 0.89 j) - (0i + 1.8j) )

= -1.57i + 0.91 j

|B - A| = sqrt(1.57^2 + 0.91^2) = 1.81 m

direction : theta = tan^-1(0.91/-1.57) = 150 degrees

D) A - B = (1.57i + 0.89 j) - 2(0i + 1.8j)

= 1.57i -2.71 j

|A-2B| = sqrt(1.57^2 + 2.71^2) = 3.13 m

direction : theta = tan^-1(-2.71/1.57) = 305.9 degrees

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.