Consider the three displacement vectors = ( 4î − 3ĵ) m, = (3î − 6ĵ) m, and = (−6î + 5ĵ) m. Use the component method to determine the following. (Take the +x direction to be to the right.)

(a) the magnitude and direction of the vector = Darrowbold = A with arrow + B with arrow + C with arrow

magnitude= m

direction= ° counterclockwise from the +x axis(b) the magnitude and direction of E with arrow = −A with arrow − B with arrow + C with arrow magnitude

Question

02-03-2020
Physics

contestada

Consider the three displacement vectors = ( 4î − 3ĵ) m, = (3î − 6ĵ) m, and = (−6î + 5ĵ) m. Use the component method to determine the following. (Take the +x direction to be to the right.)

(a) the magnitude and direction of the vector = Darrowbold = A with arrow + B with arrow + C with arrow

magnitude= m

direction= ° counterclockwise from the +x axis(b) the magnitude and direction of E with arrow = −A with arrow − B with arrow + C with arrow magnitude

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CarliReifsteck CarliReifsteck · Answer 1 · 2020-03-05T12:07:19+01:00

Answer:

(a).The magnitude and direction of the vector is 4.12 m and 284°

(b). The magnitude and direction of the vector is 19.10 m and 313°

Explanation:

Given that,

The three displacement are

[tex]A=(4i-3j)\ m[/tex]

[tex]B=(3i-6j)\ m[/tex]

[tex]C=(-6i+5j)\ m[/tex]

We need to calculate the magnitude of the vector

[tex]\vec{D}=\vec{A}+\vec{B}+\vec{C}[/tex]

Put the value into the formula

[tex]\vec{D}=(4i-3j)+(3i-6j)+(-6i+5j)[/tex]

[tex]\vec{D}=(i-4j)[/tex]

[tex]|\vec{D}|=\sqrt{(1)^2+(4)^2}[/tex]

[tex]|\vec{D}|=\sqrt{17}[/tex]

[tex]|\vec{D}|=4.12\ m[/tex]

We need to calculate the direction of the vector

Using formula of direction

[tex]\tan\theta=\dfrac{j}{i}[/tex]

[tex]\theta=\tan^{-1}(\dfrac{j}{i})[/tex]

Put the value into the formula

[tex]\theta=\tan^{-1}(\dfrac{-4}{1})[/tex]

[tex]\theta=360^{\circ}-76^{\circ}[/tex]

[tex]\theta=284^{\circ}[/tex]

(b). We need to calculate the magnitude of the vector

[tex]\vec{D}=-\vec{A}-\vec{B}+\vec{C}[/tex]

Put the value into the formula

[tex]\vec{D}=-(4i-3j)-(3i-6j)+(-6i+5j)[/tex]

[tex]\vec{D}=(-13i+14j)[/tex]

[tex]|\vec{D}|=\sqrt{(13)^2+(14)^2}[/tex]

[tex]|\vec{D}|=19.10\ m[/tex]

We need to calculate the direction of the vector

Using formula of direction

[tex]\tan\theta=\dfrac{j}{i}[/tex]

[tex]\theta=\tan^{-1}(\dfrac{j}{i})[/tex]

Put the value into the formula

[tex]\theta=\tan^{-1}(\dfrac{14}{-13})[/tex]

[tex]\theta=360^{\circ}-47^{\circ}[/tex]

[tex]\theta=313^{\circ}[/tex]

Hence, (a).The magnitude and direction of the vector is 4.12 m and 284°

(b). The magnitude and direction of the vector is 19.10 m and 313°