Q64 P - University Physics with Modern Physics

Question

A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, next 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of the starting point (Fig. P1.64). Find the magnitude and direction of the third leg of the journey. Draw the vector-addition diagram and show that it is in qualitative agreement with your numerical solution.

Step-by-Step Solution

Answer

The magnitude of the third leg of the journey is, $C=2.81km$.

The direction of the third leg of the journey is north-east.

Identification of given data

• $\overrightarrow{A}=2.00km, 0° of east , \overrightarrow{B}=3.50m, 45° south of east$
  
• The resultant vector of the sailor is $\overrightarrow{R}=5.80m$,$0° east$

Concept of resultant vector

The outcome is a sum total of vector. When you combine two or more vectors, you obtain a new one.

The resultant vector can be given as,

$R=\sqrt{A^2+B^2}$

Here, $A$and $B$ are displacement vectors.

Determine the magnitude and direction of the third leg of journey

The magnitude of third leg of journey can be evaluated as

The magnitude of third leg journey in x-direction,

$C_x=R_x-A_x-B_x$

$C_x=\left(5.80km\right)-\left(2.00km\right)-\left(3.50km\right)\left(\cos 45°\right)$

$C_x=1.33km$

The magnitude of third leg of journey in y-direction

$C_y=R_y-A_y-B_y$

$C_y=\left(0km\right)-\left(0km\right)-\left(-3.50km\right)\left(\sin 45°\right)$

$C_y=2.47km$

The resultant magnitude of third leg of journey is,

$C=\sqrt{{\left(C_x\right)}^2+{\left(C\right)}^{}}$_y2

$C=\sqrt{{\left(1.33km\right)}^2+{\left(2.47km\right)}^2}$

$C=2.81km$

Thus, the magnitude of the third leg is .

The direction of the third leg can be evaluated as,

$\theta ={\tan }^{-1}\left(\frac{2.47km}{1.33km}\right)$

$\theta =61.7$north-east

Thus, the third leg lies in the first quadrant.