Question

A person walks 34 m East and then walks 36 m at an angle 34◦ North of East.
What is the magnitude of the total dis- placement?
Answer in units of m

Answer 1

Answer on Question #46580 – Physics – Mechanics | Kinematics | Dynamics

A person walks $34\,\mathrm{m}$ East and then walks $36\,\mathrm{m}$ at an angle $34{}^\circ$ North of East.

What is the magnitude of the total displacement?

Answer in units of m

Solution:

We have two displacements: $r_1$ (when person walks 34 m East), $r_2$ (when person walks 36 m at an angle $34{}^\circ$) and total displacement $r$.

Displacement along the X-axis:

$$\begin{array}{l} r_{1x} = 34\,\mathrm{m} \\ r_{2x} = 36\,\mathrm{m} \cdot \cos(34{}^\circ) = 29.9\,\mathrm{m} \\ r_x = r_{1x} + r_{2x} = 34\,\mathrm{m} + 29.9\,\mathrm{m} = 63.9\,\mathrm{m} \\ \end{array}$$

Displacement along the Y-axis:

$$\begin{array}{l} r_{1y} = 0 \\ r_{2y} = 36\,\mathrm{m} \cdot \sin(34{}^\circ) = 20.13\,\mathrm{m} \\ r_y = r_{1y} + r_{2y} = 0 + 20.13\,\mathrm{m} = 20.13\,\mathrm{m} \\ \end{array}$$

Using the Pythagorean Theorem:

$$r^2 = r_y^2 + r_x^2$$ $$D = \sqrt{r_y^2 + r_x^2} = \sqrt{(63.9\,\mathrm{m})^2 + (20.13\,\mathrm{m})^2} = 67\,\mathrm{m}$$

Answer: the magnitude of total displacement is equal to $67\,\mathrm{m}$.

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