a 1520 kg car in motion rolls down a frictionless hill that is inclined at 14.4°. if the hill is 28.5 m…

a 1520 kg car in motion rolls down a frictionless hill that is inclined at 14.4°. 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 m/s² v_f = ? m/s

a 1520 kg car in motion rolls down a frictionless hill that is inclined at 14.4°. 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 m/s² v_f = ? m/s

Answer

Explanation:

Step1: Identify the kinematic - equation

We use the kinematic equation $v_f^2 = v_i^2+2ax$. Since the car starts from rest, $v_i = 0$.

Step2: Substitute the values

We know that $a = 2.44\ m/s^2$ and $x = 28.5\ m$. Substituting into the equation $v_f^2=0 + 2\times2.44\times28.5$. First, calculate $2\times2.44\times28.5$: $2\times2.44\times28.5=2\times69.54 = 139.08$. So, $v_f^2=139.08$.

Step3: Solve for $v_f$

Take the square - root of both sides: $v_f=\sqrt{139.08}\approx11.8\ m/s$.

Answer:

$11.8$