Find the resulting displacement of a 15-km displacement and a 25-km displacement when the angle between them is 90 degrees.( Your answer should include magnitude and direction of the displacement.)
To find the resultant, separate each of the vectors into x and y components so it should be:
v1: 15x + 0y
v2: 0x + 25 y
add those together and the resultant vector is:
resultant vector: 15x + 25y
to get the magnitude, you do the square root of the x-comp squared plus the y-comp squared so its
15^2+25^2 and then square root that and you should get a magnitude of about 29 km
for the direction of displacement, or the angle, you do tan(angle)=ycomp/xcomp so angle=inversetan(25/15) and you should get about 59 degrees north of east
To find the resulting displacement, we can use the concept of vector addition. When two vectors are at right angles to each other, we can use the Pythagorean theorem to calculate the magnitude of the resulting displacement and trigonometry to find the direction.
Given a 15-km displacement and a 25-km displacement at a right angle (90 degrees) to each other, we can visualize these vectors forming a right triangle.
To find the magnitude of the resulting displacement:
1. Square the magnitudes of the two displacements: (15 km)² + (25 km)² = 225 km² + 625 km² = 850 km²
2. Take the square root of the sum of squares: √(850 km²) ≈ 29.15 km
To find the direction of the resulting displacement:
1. Use trigonometry to find the angle. We can use the inverse tangent (tan⁻¹) function to find the angle since the displacement vectors form a right triangle.
- Take the inverse tangent of the smaller side (15 km) divided by the larger side (25 km): tan⁻¹(15 km / 25 km) ≈ 32.98 degrees
- The angle between the 15-km and 25-km displacements is approximately 32.98 degrees.
Therefore, the resulting displacement has a magnitude of approximately 29.15 km and a direction of approximately 32.98 degrees relative to one of the original displacements.