Question

a stock car accelerates uniformly from rest at 9.80m/s² for 404 m, what is his final speed?

Explanation:

Step1: Identify the kinematic equation

The kinematic equation \(v^{2}=v_{0}^{2}+2ax\) is suitable. Here, \(v_{0} = 0\) (starts from rest), \(a=9.80\ m/s^{2}\), and \(x = 404\ m\).

Step2: Substitute the values into the equation

Substitute \(v_{0}=0\), \(a = 9.80\ m/s^{2}\), and \(x=404\ m\) into \(v^{2}=v_{0}^{2}+2ax\). We get \(v^{2}=0 + 2\times9.80\times404\).
First, calculate \(2\times9.80\times404=2\times9.80\times404 = 7918.4\).
Then, find \(v\) by taking the square - root of \(7918.4\). So \(v=\sqrt{7918.4}\).
\(v\approx89.0\ m/s\)

Answer:

The final speed of the stock - car is approximately \(89.0\ m/s\)