If he travel 7km 40° west northern 10km east find the ressultant
Not sure what direction "west northern" is, but I'll assume 40° west of north. That would make the x-y components
(-7sin40°,7cos40°) = (-4.50,5.36)
10 east = (10,0)
The resultant distance is √(5.50^2 + 5.36^2) = 7.68
The angle θ from due east is
tanθ = y/x = 5.36/5.50 = 0.97
θ = 44° so the direction is
E44°N or N46°E
Need calaculation step by step
To find the resultant of the two displacement vectors, you can break them down into their horizontal and vertical components and then add them up.
First, let's break down the given displacement vectors:
Vector 1: Traveling 7 km at 40° west of north
Horizontal component: 7 km * sin(40°) (negative because it's west)
Vertical component: 7 km * cos(40°) (positive because it's north)
Vector 2: Traveling 10 km east
Horizontal component: 10 km (positive because it's east)
Vertical component: 0 km (no vertical displacement as it's purely horizontal)
Now, add up the horizontal and vertical components separately:
Horizontal component (west is negative, east is positive):
7 km * sin(40°) + 10 km = -1.92 km + 10 km = 8.08 km
Vertical component (north is positive):
7 km * cos(40°) + 0 km = 5.35 km + 0 km = 5.35 km
Finally, use the Pythagorean theorem to find the magnitude of the resultant:
Magnitude of the resultant = √(horizontal component^2 + vertical component^2)
= √(8.08 km^2 + 5.35 km^2)
= √(65.4464 km^2)
≈ 8.10 km
Therefore, the resultant of traveling 7 km 40° west of north and 10 km east is approximately 8.10 km.