Question

An rv travels 45 km east and stays the night at a KOA. The next day it travels for 3 hours to the north l, traveling 110 km. What is the displacement over the two days for the RV?

Answer 1

Answer:

The displacement of RV for the two days is 118.85 km at an angle of 67.75° with the east direction.

Explanation:

Given:

Distance moved in the East direction (d) = 45 km

Distance moved in the North direction (D) = 110 km

Displacement is defined as the difference of final position and initial position.

Let us draw a diagram representing the above situation.

Point A is the starting point and point C is the final position of RV.

So, the displacement of RV in two days is given as:

Displacement = Final position - Initial position = AC

Now, triangle ABC is a right angled triangle with AB = 45 km, BC = 110 km, and AC being the hypotenuse.

Using Pythagoras theorem, we have:

$AC^2=AB^2+BC^2\\\\AC=\sqrt{AB^2+BC^2}$

Plug in the given values and solve for AC. This gives,

$AC=\sqrt{(45\ km)^2+(110\ km)^2}\\\\AC=\sqrt{2025+12100}\ km\\\\AC=\sqrt{14125}\ km\\\\AC=118.85\ km$

Now, the direction of displacement with the east direction is given as:

$\theta=\tan^{-1}(\frac{BC}{AB})\\\\\theta =\tan^{-1}(\frac{110}{45})=67.75^\circ$

Therefore, the displacement of RV for the two days is 118.85 km at an angle of 67.75° with the east direction.