https://file.aiquickdraw.com/mathbot/file/17268542041953wzltb4r.jpg

The problem in the image asks to find the total displacement of a car that travels 35 km north and then 20 km east. Displacement is the straight-line distance from the starting point to the ending point, which can be calculated using the Pythagorean theorem, since the path forms a right triangle.

Using the Pythagorean theorem:
\[
\text{Displacement} = \sqrt{(35^2 + 20^2)}
\]
\[
\text{Displacement} = \sqrt{(1225 + 400)} = \sqrt{1625}
\]
\[
\text{Displacement} \approx 40.3 \, \text{km}
\]

Thus, the total displacement to the nearest tenth is approximately **40.3 km**, so the correct answer is **D**.

Would you like more details or have any questions?

Here are 5 related questions to expand on the topic:
1. How does displacement differ from distance in physics?
2. What are the conditions when displacement equals the distance traveled?
3. How do you calculate displacement in three dimensions?
4. Can displacement ever be negative? If so, under what circumstances?
5. What is the significance of vector components in calculating displacement?

**Tip:** Displacement considers only the straight-line distance and direction, while distance measures the total path traveled.

What is the total displacement, to the nearest tenth, of a car that travels 35 km north, then 20 km east?

Learn how to calculate the total displacement of a car that travels 35 km north and 20 km east using the Pythagorean theorem. This physics problem is suitable for students in Grades 8-10.

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