Question

A person walks 29 m East and then walks 38 m at an angle 45◦ North of East.
What is the magnitude of the total displacement?

Answer 1

Question #45259 - Physics - Mechanics | Kinematics | Dynamics

A person walks $29\,\mathrm{m}$ East and then walks $38\,\mathrm{m}$ at an angle $45{}^{\circ}$ North of East.

What is the magnitude of the total displacement?

Solution:

We have two displacements: $r_1$ (when person walks 29 m East), $r_2$ (when person walks 38 m at an angle $45{}^\circ$) and total displacement $r$.

Displacement along the X-axis:

$$\begin{array}{l} r_{1x} = 29\,\mathrm{m} \\ r_{2x} = 38\,\mathrm{m} \cdot \cos(45{}^\circ) = 26.9\,\mathrm{m} \\ r_x = r_{1x} + r_{2x} = 29\,\mathrm{m} + 26.6\,\mathrm{m} = 55.6\,\mathrm{m} \\ \end{array}$$

Displacement along the Y-axis:

$$\begin{array}{l} r_{1y} = 0 \\ r_{2y} = 38\,\mathrm{m} \cdot \sin(45{}^\circ) = 26.9\,\mathrm{m} \\ r_y = r_{1y} + r_{2y} = 0 + 26.9\,\mathrm{m} = 26.9\,\mathrm{m} \\ \end{array}$$

Using the Pythagorean Theorem:

$$r^2 = r_y^2 + r_x^2$$ $$D = \sqrt{r_y^2 + r_x^2} = \sqrt{(55.6\,\mathrm{m})^2 + (26.9\,\mathrm{m})^2} = 61.8\,\mathrm{m}$$

Answer: the magnitude of total displacement is equal to $61.8\,\mathrm{m}$.

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