Question

a 1520 kg car in motion rolls down a frictionless hill that is inclined at \\(14.4^{\circ}\\).

if the hill is 28.5 m long, what is the velocity of the car as it reaches the base of the hill?

\\a = 2.44\text{ m/s}^2\\
\\v_{\text{f}} = ?\text{ m/s}\\

Explanation:

🆕 New Concept Discovered: Conservation of Energy or Kinematics on an Incline
Using motion equations to find final velocity.

Step 1: Identify given values

The problem provides the following values:

• Mass of the car, \( m = 1520\text{ kg} \) (not needed for kinematics)
• Acceleration down the incline, \( a = 2.44\text{ m/s}^2 \)
• Distance along the incline, \( d = 28.5\text{ m} \)
• Initial velocity, \( v_i = 0\text{ m/s} \) (since it "rolls down" starting from rest)

Step 2: Select the kinematic equation

To find the final velocity \( v_f \) when acceleration \( a \), distance \( d \), and initial velocity \( v_i \) are known, we use:

$$ v_f^2 = v_i^2 + 2ad $$

Step 3: Calculate the final velocity

Substitute the known values into the equation:

$$ v_f^2 = 0^2 + 2(2.44\text{ m/s}^2)(28.5\text{ m}) $$

$$ v_f^2 = 139.08 $$

$$ v_f = \sqrt{139.08} \approx 11.79\text{ m/s} $$

Rounding to three significant figures gives:

$$ v_f \approx 11.8\text{ m/s} $$

Answer:

11.8