A student walked 4 km due east and then 15km in a direction of 37° north of east ,finally the walked due north for 3km ,how much is this displacement?
East:
E = 4 + 15 cos 37
North:
N = 3 + 15 sin 37
d = sqrt (E^2+N^2)
angle A clockwise from north
tan A = E / N
0972209372
18.4km
20km
To calculate the displacement, we need to find the total distance traveled and the direction of the final displacement.
Step 1: Calculate the displacement along the east-west direction.
The student walked 4 km due east. This means the displacement along the east-west direction is 4 km east.
Step 2: Calculate the displacement along the northeast direction.
The student walked 15 km in a direction 37° north of east. To find the displacement along this direction, we need to find the east and north components separately using trigonometry.
East component = 15 km * cos(37°)
North component = 15 km * sin(37°)
Step 3: Calculate the displacement along the north-south direction.
The student walked 3 km due north. This means the displacement along the north-south direction is 3 km north.
Step 4: Add up all the displacements to get the total displacement.
Total east-west displacement = 4 km east
Total north-south displacement = (East component) + (North component) + 3 km north
To calculate the displacement, you can use the Pythagorean theorem:
Displacement = √((Total east-west displacement)² + (Total north-south displacement)²)
Substituting the values:
Displacement = √((4 km)² + ((East component) + (North component) + 3 km)²)
Now let's substitute the values and solve.
East component = 15 km * cos(37°) ≈ 12.017 km
North component = 15 km * sin(37°) ≈ 8.973 km
Displacement = √((4 km)² + (12.017 km + 8.973 km + 3 km)²)
Displacement = √((4 km)² + (24.99 km)²)
Displacement ≈ √(16 km² + 624.01 km²)
Displacement ≈ √640.01 km²
Displacement ≈ 25.2 km
Therefore, the displacement is approximately 25.2 km.