Question

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?

Answer 1

Answer:12.206 cm,$\theta =54.99^{\circ}$

Explanation:

Given

Insect walks 15 cm to the right

so its position vector is$r_1=15i$

Now it moves 10 cm up so its new position vector

$r_2=15i+10j$

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

$r_3=15\hat{i}+10\hat{j}-8\hat{i}=7\hat{i}+10\hat{j}$

so its displacement is given by

$|r_3|=\sqrt{7^2+10^2}=\sqrt{149}=12.206 cm$

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

$tan\theta =\frac{10}{7}=1.428$

$\theta =54.99^{\circ}$