Q33E - University Physics with Modern Physics

Question

A disoriented physics professor drives 3.25 km north, then 2.20 km west, and then 1.50 km south. Find the magnitude and direction of the resultant displacement, using the method of components. In a vector-addition diagram (roughly to scale), show that the resultant displacement found from your diagram is in qualitative agreement with the result you obtained by using the method of components.

Step-by-Step Solution

Answer

The total displacement was $2.811$km and the direction of the vector was southeast at an angle of  $141.5^o$

Vector and its direction

The vector's various coordinates are presented here. We know that in coordinates, the x component of a vector comes first, followed by the y component. To calculate the angle of the vector with respect to the x-axis, just multiply the y component by the x component and then take the inverse of that result.

This will tell you the vector's angle with the x-axis

If we draw rough diagram of the path of the professor then it would be as below

Rough path of the professor

The above figure describes the complete path of the professor

Here total three times he is changing his position so there will be three vectors.

Let's name these vectors as A,B and C. 

So from the figure it is clear that

$$\begin{gathered} A=(0 \mathrm{km},-3.25 \mathrm{km}) \\ B=(-2.2 \mathrm{km},0 \mathrm{km}) \\ C=(0 \mathrm{km},-1.5 \mathrm{km}) \end{gathered}$$

Resultant displacement vector and magnitude

Therefore the resultant vector will be the sum of all these three vectors.

$$\begin{gathered} R=A+B+C \\ =\left[\right(0 \mathrm{km},-3.25 \mathrm{km})+(-2.2 \mathrm{km},0 \mathrm{km})+(0 \mathrm{km},-1.5 \mathrm{km}\left)\right] \\ =(-2.2 \mathrm{km},1.75 \mathrm{km}) \end{gathered}$$

Hence the resultant displacement vector will be $(-2.2 \mathrm{km},1.75 \mathrm{km})$

Magnitude of the Displacement vector

The magnitude of the vector can be calculated by the sum of the square of the magnitude of the individual coordinates

$$\begin{gathered} A=\sqrt{{\left(A_x\right)}^2+{\left(B_x\right)}^2} \\ =\sqrt{(-2.2 \mathrm{km}{)}^2+(1.75 \mathrm{km}{)}^2} \\ =2.811 \mathrm{km} \end{gathered}$$

Hence the magnitude of the vector is $2.811$ kilometers.

Direction of the Displacement vector

The direction of the vector will be the ratio of the y coordinates with that of the ratio of the $x$ coordinate.

For our case$A_x=-2.2$ and $A_y=1.75$ kilometers

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{A_y}{A_x}\right) \\ ={\tan }^{-1}\left(-\frac{1.75 \mathrm{km}}{2.2 \mathrm{km}}\right) \\ =180°-38.5° \\ =141.5° \end{gathered}$$

measured counterclockwise from the positive $x$-axis.