Answer to Question #179357 in Classical Mechanics for Kristen Tyler

Question

A car slows from 23.69 m/s to rest in 4.44 s. How far did it travel in this time?

Answer 1

Answer

Using

v=u+at

$0=23.69-a\times4.44\\a=5.35m/s^2$

So travelled distance

$S=\frac{v^2-u^2}{2a}\\=\frac{0-23.69^2}{-2\times5.35}\\=52.45m$