Question

A student leaves a sidewalk corner walks 3.0 blocks north then walks 3.0 blocks west.what is the students displacement from the starting position

Answer 1

Answer:

The students displacement is 3·√3 blocks, 45° North of West

Explanation:

From the question, we have;

The distance North in which the student walks = 3.0 blocks

The next walking distance West by the student = 3.0 blocks

The displacement of the student = The shortest distance between the student start and final point

By representing the motion of the student in vector format, we have;

$\mathbf{d} = 3\cdot \hat{i} + 3\cdot \hat{{j}}$

Therefore the magnitude of d is given as follows;

$\left | d\right | = \sqrt{3^2 + 3^2} = 3\cdot \sqrt{3} \ blocks$

The direction of the student = tan⁻¹(3/3) = 45° West of North

The students displacement = 3·√3 blocks, 45° North of West.