The displacement of the hiker can be found by treating the eastward and northward movements as vectors. Using the Pythagorean theorem, the displacement is the square root of the sum of the squares of the distances traveled in each direction. In this case, the displacement is √(3.0 km)^2 + (4.0 km)^2 = √(9 km^2 + 16 km^2) = √25 km^2 = 5.0 km. Therefore, the displacement of the hiker is 5.0 km in a direction that is 53.1 degrees north of east.
What is the displacement of a car traveling 10 km north 5km east 15 km south and 5 km north?
20ms east
From the information that you provided, it is not possible to determine the displacement of the car. However, I can answer your second question. The distance is: 15+10+25+10=60 km
That is incorrect. The distance travelled north cancels out the distance travelled south. Therefore - he only travels three blocks east.
3.00 m/s
1 mile East
a hiker walks 15 mi due east, then heads due north for 8 km. what is the direction of the resultant vector?
The resulting displacement can be found by using the Pythagorean theorem: (9 miles)^2 + (7 miles)^2 = c^2 81 + 49 = c^2 c = √130 Therefore, the resulting displacement is approximately 11.4 miles in a northeast direction.
The cyclist ends up 1 mile east of the starting point (unless the cycling takes place near the north or south pole!). So the displacement is 1 mile in an easterly direction.
5km east
The displacement of the person pacing back and forth travels 5 west 7m west is usually East.
What is the displacement of a car traveling 10 km north 5km east 15 km south and 5 km north?
What is the displacement of a car traveling 10 km north 5km east 15 km south and 5 km north?
20ms east
6.3 meter
Type your answer here... C.3.5 km at 50º north of west
find the resultant of the following displacement a=20km 30south of east