Final answer:
The hovercraft's initial velocity components are -2.828 m/s to the west and 2.828 m/s to the north. The acceleration components are 10 m/s² to the east and 0 m/s² to the north. The final velocity is 27.3 m/s at an angle of 5.9° north of east.
Step-by-step explanation:
The student question involves a hovercraft with initial and final velocities and acceleration. Let's address each part separately:
a. Initial Velocity Components
The hovercraft has an initial velocity of 4 m/s northwest. To find the components, we need to split this vector into its x (west to east) and y (south to north) components. Northwest implies a 45-degree angle to both the horizontal and vertical axes, thus:
Vx_initial = V_initial * cos(45°) = 4 m/s * cos(45°) = 4 m/s * (\(√{2}/2\)) ≈ -2.828 m/s (negative due to westward direction)
Vy_initial = V_initial * sin(45°) = 4 m/s * sin(45°) = 4 m/s * (\(√{2}/2\)) ≈ 2.828 m/s (positive due to northward direction)
b. Acceleration Components
The hovercraft accelerates to the east with an acceleration of 10 m/s². Since the acceleration is purely horizontal (eastward), the components are:
Ax = 10 m/s²
Ay = 0 m/s²
c. Final Velocity Components
The final velocity components can be calculated using the formula V_final = V_initial + a*t:
Vx_final = Vx_initial + Ax*t = -2.828 m/s + 10 m/s² * 3 s = 27.172 m/s
Vy_final = Vy_initial + Ay*t = 2.828 m/s + 0 m/s² * 3 s = 2.828 m/s
d. Magnitude and Direction of Final Velocity
The magnitude of the final velocity can be calculated using the Pythagorean theorem:
|V_final| = \(√{(Vx_final)^2 + (Vy_final)^2)}\) = \(√{(27.172 m/s)^2 + (2.828 m/s)^2)}\) ≈ 27.3 m/s
To find the direction, we use arctan(Vy_final/Vx_final):
θ = arctan(2.828/27.172) ≈ arctan(0.104) ≈ 5.9° north of east.