A car travels 10 km southeast and then 15 km in a direction 60° north of east. Find the magnitude and direction of the cars resultant vector. (Round to the nearest tenth)

Answer 1

A car travels 10km SE, then 15km 60° North of east. Find the magnitude and direction of the cars resultant vector, rounded to the nearest 10th.

You need to draw this out! Start at the origin (0,0), go on a -45° heading into the 4th quadrant by a distance of 10. Using the trig functions (or special triangles); you will see 10 cos -45° is the x value and 10 sin -45° is the y value. (x₁, y₁) = (10/√2, - 10√2) ≅ (7.07, -7.07). From that point, go a distance of 15 at 60°. Use trig function again to find the next/last coordinate. The x value to add to the first coordinate would be + 15 cos 60°, and the y value to add to the first coordinate would be + 15 sin 60°); so (x₂, y₂) = (7.07°+ 15 cos 60 , -7.07 + 15 sin 60°) ≅ (7.07, -7.07) ≅ (14.57, 5.92)

From the origin to that point is the resultant vector.

1. To calculate the direction from the origin (0,0) to (x₂, y₂) use trig functions. You need to solve for θ and you have the x and y (or the opposite and adjacent legs); so use Tan θ = 5.92/14.57, and take the inverse Tan of both to solve for θ, that's the direction. θ = 22°7'
2. To calculate the magnitude, just use the distance formula (0,0) to (14.57, 5.92); I will presume you can do that part.

Answer 2

-10sin45+15cos60=Hor distance = 5.92 km

-10cos45+15sin60=ver distance = .429 km

resultant is sqrt(5.92²+.429²)= 5.94 km at angle arctan(.429/5.92)= 1.77 degrees north of east