Answer to Question #294318 in Physics for mrymmm

Question

A car is traveling to the left, which is the negative direction. The direction of travel remains the same throughout this problem. The car's initial speed is 34.1 m/s, and during a 5.87-second interval, it changes to a final speed of (a)51.1 m/s and (b)30.6 m/s. In each case, find the acceleration (magnitude and algebraic sign)

Answer 1

Given:

$v_{ix}=-34.1\:\rm m/s$

$t=5.87\:\rm s$

The acceleration of the car

$a_x=\frac{v_{fx}-v_{ix}}{t}$

(a)

$a_x=\frac{-51.1-(-34.1)}{5.87}=-2.90\:\rm m/s^2$

(b)

$a_x=\frac{-30.6-(-34.1)}{5.87}=0.60\:\rm m/s^2$