Question

Starting from rest, a car accelerates at 2.4 m/s2 up a hill that is inclined 5.5° above the horizontal. how far horizontally and vertically has the car traveled in 12 s?

Answer 1

To calculate the horizontal and vertical distances traveled by the car, use kinematic equations incorporating the angle of the hill's incline. The horizontal distance is approximately 172 meters and the vertical distance is approximately 15.9 meters after 12 seconds of acceleration.

Step-by-step explanation:

Calculating Distance of a Car Accelerating Up a Hill

When a car starting from rest accelerates up a hill at 2.4 m/s² for 12 seconds and the hill is inclined at 5.5 degrees above the horizontal, we will calculate the horizontal and vertical distances traveled using the kinematic equations of motion. The scenario is a physics problem involving kinematics and components of motion along a slope.

Known variables:

Acceleration (a) = 2.4 m/s²

Time (t) = 12 s

Angle of incline (Θ) = 5.5°

Horizontal distance (x): We use the equation x = v₀ t + (1/2) a t² cos(Θ), where v₀ is the initial velocity which is 0.

x = 0 + (1/2) × 2.4 m/s² × (12 s)² × cos(5.5°)

x ≈ 172 m (horizontal distance)

Vertical distance (y): We use the equation y = (1/2) a t² sin(Θ).

y = (1/2) × 2.4 m/s² × (12 s)² × sin(5.5°)

y ≈ 15.9 m (vertical distance)

The units are consistent (meters for distance, seconds for time), and the result seems reasonable for a car accelerating moderately for 12 seconds.

Answer 2

The car travels 172.0m horizontally and 16.6m vertically The car travels 172.8m in 12s, given by the equation d = ½at² = ½ * 2.4m/s² * (12s)² The horizontal distance is given by x = d * cos(θ) = 172.8m * cos(5.5°) = 172.0m The vertical distance is given by y = d * sin(θ) = 172.8m * sin(5.5°) = 16.6m NOTE: when calculating cos() and sin(), make sure your calculator is in degree mode