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An insect meanders across a sidewalk. The insect moves 15cm to the right, 10cm up the sidewalk and 8cm to the left. What is the magnitude and direction of the insect's total displacement?

Respuesta :

Answer:12.206 cm,[tex]\theta =54.99^{\circ}[/tex]

Explanation:

Given

Insect walks 15 cm to the right

so its position vector is[tex]r_1=15i[/tex]

Now it moves 10 cm up so its new position vector

[tex]r_2=15i+10j[/tex]

Now it moves 8 cm left so its final position vector is

[tex]r_3=15\hat{i}+10\hat{j}-8\hat{i}=7\hat{i}+10\hat{j}[/tex]

so its displacement is given by

[tex]|r_3|=\sqrt{7^2+10^2}=\sqrt{149}=12.206 cm[/tex]

For direction, let \theta is the angle made by its position vector with x axis

[tex]tan\theta =\frac{10}{7}=1.428[/tex]

[tex]\theta =54.99^{\circ}[/tex]

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