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?

The displacement of the rv over the two days is 118.85 km.

The given parameters;

The displacement of the rv over the two days is calculated by applying Pythagoras theorem as follows;

Thus, the displacement of the rv over the two days is 118.85 km.

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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:

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

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

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

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