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ollegr [7]
4 years ago
14

Vector A points in the negative y-direction and has a magnitude of 14 units. Vector B has twice the magnitude and points in the

positive x-direction.
(a) Find the direction and magnitude of A + B. (degrees counterclockwise from the +x axis)
(b) |A + B| = ?
(c) Find the direction and magnitude of A - B. (degrees counterclockwise from the +x axis)
(d) |A - B| =?
(e) Find the direction and magnitude of B - A. (degrees counterclockwise from the +x axis)
Physics
1 answer:
Nutka1998 [239]4 years ago
4 0

Answer:

(a)

|\vec{A} + \vec{B}| = 31.3~{\rm units}\\\theta = 333.4^\circ

(b)

|\vec{A} + \vec{B}| = 31.3~{\rm units}

(c)

|\vec{A} - \vec{B}| = 31.3~{\rm units}\\\theta =206.56^\circ

(d)

|\vec{A} - \vec{B}| = 31.3~{\rm units}

(e)

|\vec{B} - \vec{A}| = 31.3~{\rm units}\\\theta =26.5^\circ

Explanation:

\vec{A} = -14\^y\\\vec{B} = 28\^x

(a)

\vec{A} + \vec{B} = 28\^x -  14\^y\\|\vec{A} + \vec{B}| = \sqrt{28^2 + (-14^2)} = 31.3~{\rm units}\\\tan(\theta) = \frac{y}{x} = -0.5\\\theta = \arctan(-0.5) = 333.4^\circ

Here, the vector is in the fourth quadrant. Hence, we subtracted the arctan value from 360°, because the arctan value gives the angle from the x-axis.

(b)

|\vec{A} + \vec{B}| = \sqrt{28^2 + (-14^2)} = 31.3~{\rm units}

(c)

\vec{A} - \vec{B} = -28\^x - 14\^y\\|\vec{A} - \vec{B}| = \sqrt{(-28)^2 + (-14^2)} = 31.3~{\rm units}\\\tan(\theta) = \frac{y}{x} = 0.5\\\theta = \arctan(0.5) = 206.56^\circ

Here, the vector is in the third quadrant. Hence, we added 180° to the arctan value.

(d)

|\vec{A} - \vec{B}| = \sqrt{(-28)^2 + (-14^2)} = 31.3~{\rm units}

(e)

\vec{B} - \vec{A} = 28\^x +  14\^y\\|\vec{B} - \vec{A}| = \sqrt{28^2 + 14^2} = 31.3~{\rm units}\\\tan(\theta) = \frac{y}{x} = 0.5\\\theta = \arctan(0.5) = 26.5^\circ

Here, the vector is in the first quadrant.

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speed is maximum when sine is + -1

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