An object is given two initial velocities, one that has a magnitude of 60.0 ms-1 and points along a direction that is 35.0° north of east and the other has a magnitude of 55.0 ms-1 that points 15.0°, also north of east. Determine the resultant velocity of the object.

do this with north (y) and east (x) components the same way as the one you asked below.

Use the formula for resultant of two vectors.

R = √(3600 + 3025 + 2*60*55*cos20)

You should also find direction of resultant using the tan fromula

To determine the resultant velocity of the object, we need to find the vector sum of the two given velocities.

Step 1: Convert the given velocities into their x and y components.

For the first velocity:
Magnitude = 60.0 m/s
Angle = 35.0° north of east

To find the x-component:
V₁x = Magnitude * cos(angle)

V₁x = 60.0 m/s * cos(35.0°)

V₁x ≈ 49.144 m/s

To find the y-component:
V₁y = Magnitude * sin(angle)

V₁y = 60.0 m/s * sin(35.0°)

V₁y ≈ 34.268 m/s

For the second velocity:
Magnitude = 55.0 m/s
Angle = 15.0° north of east

To find the x-component:
V₂x = Magnitude * cos(angle)

V₂x = 55.0 m/s * cos(15.0°)

V₂x ≈ 52.951 m/s

To find the y-component:
V₂y = Magnitude * sin(angle)

V₂y = 55.0 m/s * sin(15.0°)

V₂y ≈ 14.978 m/s

Step 2: Add the x-components and y-components separately.

Vx = V₁x + V₂x

Vx ≈ 49.144 m/s + 52.951 m/s

Vx ≈ 102.095 m/s

Vy = V₁y + V₂y

Vy ≈ 34.268 m/s + 14.978 m/s

Vy ≈ 49.246 m/s

Step 3: Combine the x and y components to find the magnitude and direction of the resultant velocity.

To find the magnitude of the resultant velocity:
Resultant Velocity = sqrt(Vx² + Vy²)

Resultant Velocity ≈ sqrt((102.095 m/s)² + (49.246 m/s)²)

Resultant Velocity ≈ sqrt(10421.15 m²/s² + 2425.24 m²/s²)

Resultant Velocity ≈ sqrt(12846.39 m²/s²)

Resultant Velocity ≈ 113.42 m/s

To find the direction of the resultant velocity:
Direction = tan^(-1)(Vy / Vx)

Direction ≈ tan^(-1)(49.246 m/s / 102.095 m/s)

Direction ≈ tan^(-1)(0.4814)

Direction ≈ 25.42°

Therefore, the resultant velocity of the object is approximately 113.42 m/s at an angle of 25.42° north of east.