Q67P - University Physics with Modern Physics

Question

As a test of orienteering skills, your physics class holds a contest in a large, open field. Each contestant is told to travel 20.8 m due north from the starting point, then 38.0 m due east, and finally 18.0 m in the direction 33.0^° west of south. After the specified displacements, a contestant will find a silver dollar hidden under a rock. The winner is the person who takes the shortest time to reach the location of the silver dollar. Remembering what you learned in class, you run on a straight line from the starting point to the hidden coin. How far and in what direction do you run?

Step-by-Step Solution

Answer

The magnitude that the winner will run is,$R=28.8\mathrm{m}$$R=28.8\mathrm{m}$.

The direction that the winner will run is,${\theta }_R=11.4°\mathrm{north}-\mathrm{east}$.

Step 1: Identification of given data

• The magnitude of the winner is in the A direction is,$A=20.8\mathrm{m}$
• The magnitude of the winner is in the B direction is,$B=38.0\mathrm{m}$
• The magnitude of the winner is in the C direction is,$C=18.0\mathrm{m}$

Step 2: Concept of displacement

Displacement is defined as straight line distance covered in a certain direction.

A displacement vector is defined as the longest vector whose length equals the smallest distance between the beginning and end positions.

Step 3: Determine the distance travelled by the winner and direction of the winner

The diagram has been provided below-

The equation of the components of magnitude of the vectors A and B is the x and y direction are given by, 

$$\begin{gathered} A_x=A\cos 90°, \\ {A}_y=A\sin 90° \\ {B}_x=B\cos 0° \\ {B}_y=B\sin 0° \end{gathered}$$

Here,${\mathrm{A}}_x\mathrm{and} {\mathrm{A}}_{\mathrm{Y}}$are the x and y components and the vector A andare the x and y components and the vector B

 

Substituting the values in the above equation, 

 $$\begin{gathered} {\mathrm{A}}_{\mathrm{x}}=0, \\ {\mathrm{A}}_{\mathrm{y}}=20.8\mathrm{m}, \\ {\mathrm{B}}_{\mathrm{x}}=38.0\mathrm{m}, \\ {\mathrm{B}}_{\mathrm{y}}=0 \end{gathered}$$

 

The equation of the components of magnitude of the vectors C is the x and y direction are given by, 

 $$\begin{gathered} C_x=-C\sin 33°, \\ {C}_y=-C\cos 33°, \end{gathered}$$ 

Here,$C_{\mathrm{x}} \mathrm{and} {\mathrm{C}}_{\mathrm{y}}$are the x and y components and the vector C

 

Substituting the values in the above equation,

$$\begin{gathered} C_x=-\left(18\mathrm{m}\right)\left(\sin 33.0°\right) \\ =-9.80 \mathrm{m} \\ {\mathrm{C}}_{\mathrm{y}}=-\left(18\mathrm{m}\right)\left(\cos 33.0°\right) \\ =-15.09 \mathrm{m} \end{gathered}$$

The resultant magnitude of the winner in the x direction is given by,

 $R_x=A_x+B_x+C_x$

Substituting the values in the above equation,

 

$$\begin{gathered} R_x=0+38.0\mathrm{m}+\left(-9.80\mathrm{m}\right) \\ =28.2\mathrm{m} \end{gathered}$$

 

The resultant magnitude of the winner in the y direction is given by,

 $R_y=A_y+B_y+C_y$

 

Substituting the values in the above equation,

 

$$\begin{gathered} R_y=\left(20.8\mathrm{m}\right)+0+\left(-15.09\mathrm{m}\right) \\ =5.7 \mathrm{m} \end{gathered}$$

The equation of the resultant vector is expressed as,

 

$R=\sqrt{R_x^2+R_y^2 }$

 

Here, R is the resultant vector

 

Substituting the values in the above equation,

$$\begin{gathered} R=\sqrt{{\left(28.2\mathrm{m}\right)}^2+\left(5.7\mathrm{m}\right) } \\ =28.8 \mathrm{m} \end{gathered}$$

 

Thus, the winner will have to travel$R=28.8\mathrm{m}$to cover shortest distance.

 

The equation of the direction of the winner is given by,

$\tan \theta =\frac{R_y}{R_x}$

Here,$\tan \theta$is the angle of the direction of the winner

 

Substituting the values in the above equation,

 $$\begin{gathered} \tan \theta =\frac{5.70 \mathrm{km}}{28.2 \mathrm{km}} \\ \mathrm{\theta } ={\tan }^{-1}\left(\frac{5.70 \mathrm{km}}{28.2 \mathrm{km}}\right) \end{gathered}$$

Thus, the winner will travel$11.4° \mathrm{north}‐\mathrm{east}$.