Ships a and b leave port together. for the next two hours, ship a travels at 24mph in a direction 40 degree west of north while ship b travels 29 degree east of north at 32mph .

what is the distance between the ships 2 hours later?

what is the speed of ship a as seen by ship b?

We will consider the positive x-axis to be the reference point and measures angles from it.
Angle of ship A = 90 + 40 = 130°
Angle of ship B = 90 - 29 = 61°
y-component of A's velocity:
vyA = 24sin(130) = 18.39 mph
x-component of A's velocity:
vxA = 24cos(130) = -15.43 mph
Using the same for B:
vyB = 32sin(61) = 27.99 mph
vxB = 32cos(61) = 15.51 mph
Now, we find the distance covered each in the x and y directions:
syA = 18.39 x 2 = 36.78 miles
sxA = -15.43 x 2 = -30.86 miles
Using the same for B:
syB = 55.98 miles
sxB = 31.02 miles
using the distance formula:
d = √((x₂ - x₁)² + (y₂ - y₁)²
d = √((31.02 + 30.86)² + (55.98 - 36.78)²)
d = 64.79 miles

For the velocity, we use the same method because the relative velocity is obtained by finding the difference of two velocities:
Vx = vxA - vxB
= -15.43 - 15.51 = -30.94 mph
Vy = vyA - vyB
= 18.39 - 27.99 = -9.6 mph
V resultant = √(Vx² + Vy²)
VR = √(-30.94)² + (-9.6)²
VR = 32.40 mph