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)`

Step-by-step 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|=√(7^2+10^2)=√(149)=12.206 cm`

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

`tan\theta =(10)/(7)=1.428`

`\theta =54.99^(\circ)`