Question

A hiker walks due east for a distance of 25.5 km from her base camp. On the second day, she walks 41.0 km northwest till she discovers the cave she wanted to see. Determine the magnitude and direction of her resultant displacement between the base camp and the cave​

Answer 1

Resultant displacement is 29.2 km at $83.1^{\circ}$ north of west

Explanation:

To solve the problem, we have to use the rules of vector addition, resolving first each vector along the x- and y- direction.

Taking east as positive x direction and north as positive y- direction, we have:

- First displacement is 25.5 km east, therefore its components are

$A_x = 25.5 km\\A_y = 0 km$

- Second displacement is 41.0 km northwest, so its components are

$B_x = (41.0)cos(135^{\circ})=-29.0 km\\B_y =(41.0)sin(135^{\circ})=29.0 km$

So, the components of the resultant displacement are

$R_x=A_x+B_x=25.5+(-29.0)=-3.5 km\\R_y=A_y+B_y=0+29.0=29.0 km$

And so, the magnitude is calculated using Pythagorean's theorem:

$R=\sqrt{R_x^2+R_y^2}=\sqrt{(-3.5)^2+(29.0)^2}=29.2 km$

And the direction is given by

$\theta=tan^{-1}(\frac{R_y}{|R_x|})=tan^{-1}(\frac{29.0}{3.5})=83.1^{\circ}$

Where the angle is measured from the west direction, since Rx is negative.

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