A pet-store supply truck moves at 25.0 m/s north along a highway. inside, a dog moves at 1.75 m/s at an angle of 35.0° east of north. what is the velocity of the dog relative to the road

26.5 m/s at 2.17° east of north

Explanation:

The velocity component in the north direction:

vᵧ = 25.0 m/s + 1.75 m/s cos 35.0°

vᵧ = 26.4 m/s

The velocity component in the east direction:

vₓ = 1.75 m/s sin 35.0°

vₓ = 1.00 m/s

The magnitude of the velocity is:

v² = vₓ² + vᵧ²

v² = (1.00 m/s) ² + (26.4 m/s) ²

v = 26.5 m/s

The direction is:

θ = atan (vₓ / vᵧ)

θ = atan (1.00 m/s / 26.4 m/s)

θ = 2.17° east of north

15.0 m/s^2