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

Question

The displacement vectors A and B shown in the figure below both have magnitudes of 3.95 m. The direction of vector A is 36.5o. (a) Find A + B graphically magnitude direction 59.31 6.22 Your response differs from the correct answer by more than 10%. Double check your calculations. m counterclockwise from the +x axis (b) Find A - B graphically. 3.95 Your response differs from the correct answer by more than 10%. Double check your calculations. m 323.50 Your response differs from the correct answer by more than 10%. Double check your calculations.° counterclockwise from the +x axis magnitude direction (C) FindB -A graphically magnitude 3.95 direction 143.5 Your response differs from the correct answer by more than 10%. Double check your calculations. m counterclockwise from the +x axis (d) Find A - 2B graphically 4.84 Your response differs from the correct answer by more than 10%. Double check your calculations. m 251 Your response differs from the correct answer by more than 10%. Double check your calculations.° counterclockwise from the +x axis magnitude direction

Explanation / Answer

A = 3.95*cos(36.5) i + 3.95*sin(36.5)j

A = 3.18 i + 2.35j

B = 3.95j

a)

A + B = (3.18 i + 2.35j) + (3.95j) = 3.18i + 6.3j

magnitude = sqrt(3.18^2 + 6.3^2) = 7.1

angle = tan^-1(6.3/3.18) = 63 degree

b)

A -B = (3.18 i + 2.35j) - 3.95j = (3.18i - 1.6j )

magnitude = sqrt(3.18^2 + 1.6^2) = 3.6

tan^-1(-1.6 / 3.18) = 333 degree

c)

B - A = (3.95j ) - (3.18 i + 2.35j)

B = -3.18 i + 1.6 j

magnitude = sqrt(3.18^2 + 1.6^2) = 3.6

angle = tan^-1 ( 1.6 / -3.18) = 153 degree

d)

A- 2B = (3.18 i + 2.35j) - 2(3.95j) = 3.18i - 7.9j

magnitude = sqrt(3.18^2 + 7.9^2) = 5.2

angle = tan^-1(7.9/3.18) = 232 degree

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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