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2 years ago

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A drunken sailor stumbles 580 meters north, 530 meters northeast, then 480 meters northwest. What is the total displacement and

the angle of the displacement? (Assume east is the +x-direction and north is the +y-direction.)
a) What's the magnitude?
b) What's the direction? (In degrees and counterclockwise to the x-axis.)

1 answer:

kvv77 [185]2 years ago

4 0

Answer:

(a) 1294.66 m

(b) 88.44°

Explanation:

d1 = 580 m North

d2 = 530 m North east

d3 = 480 m North west

(a) Write the displacements in vector forms

$\overrightarrow{d_{1}}=580\widehat{j}$

$\overrightarrow{d_{2}}=530\left ( Cos45\widehat{i}+Sin45\widehat{j} \right )$

$\overrightarrow{d_{2}}=374.77\widehat{i}+374.77\widehat{j}$

$\overrightarrow{d_{3}}=480\left ( - Cos45\widehat{i}+Sin45\widehat{j} \right )$

$\overrightarrow{d_{3}}=-339.41\widehat{i}+339.41widehat{j}$

The resultant displacement is given by

$\overrightarrow{d}\overrightarrow{d_{1}}+\overrightarrow{d_{2}}+\overrightarrow{d_{3}}$

$\overrightarrow{d}=\left ( 374.77-339.41 \right )\widehat{i}+\left ( 580+374.77+339.41 \right )\widehat{j}$

$\overrightarrow{d}=35.36\widehat{i}+1294.18\widehat{j}$

magnitude of the displacement

$d ={\sqrt{35.36^{2}+1294.18^{2}}}=1294.66 m$

d = 1294.66 m

(b) Let θ be the angle from + X axis direction in counter clockwise

$tan\theta =\frac{1294.18}{35.36}=36.6$

θ = 88.44°

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Explanation:

From the question we are told that

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$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

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GenaCL600 [577]

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