Question

2)a car accelerates at ( a = 2.5\frac{m}{s^{2}} ) from rest ( 0% ) over a displacement of sixty four meters( (x = 64m) ). what is the cars final velocity in meters per second?

Explanation:

Step1: Identify the kinematic equation

We use the equation \(v^{2}=v_{0}^{2}+2ax\). Here, \(v_{0} = 0\ m/s\) (starts from rest), \(a=2.5\ m/s^{2}\), and \(x = 64\ m\).

Step2: Substitute the values into the equation

Substituting \(v_{0}=0\), \(a = 2.5\), and \(x=64\) into \(v^{2}=v_{0}^{2}+2ax\), we get \(v^{2}=0^{2}+2\times2.5\times64\).
First, calculate \(2\times2.5\times64\): \(2\times2.5\times64=5\times64 = 320\). So \(v^{2}=320\).

Step3: Solve for \(v\)

Take the square - root of both sides. \(v=\sqrt{320}\). Simplify \(\sqrt{320}=\sqrt{64\times5}=8\sqrt{5}\approx 17.9\ m/s\)

Answer:

The car's final velocity is approximately \(17.9\ m/s\)