Answer to Question #230421 in Physics for Yzang

Question

A car is driving at speed of 80 mph in a circular track with a radius of 1500
ft. The driver applied brake to decelerate the car and after 6 seconds, the speed
decreases to half the original speed. What is the acceleration of the car at the
instant after the brakes have been applied? At ft/s

Answer 1

Given:

$v_i=80\:\rm mph=80*1.467\:\rm ft/s=117\:\rm ft/s$

$d=1500\:\rm ft$

$v_f=40\:\rm mph=40*1.467\:\rm ft/s=59\:\rm ft/s$

The distance traveled by a car

$$d=\frac{v_f^2-v_i^2}{2a}$$

Hence, the acceleration of a car

$$a=\frac{v_f^2-v_i^2}{2d}$$

$$a=\frac{59^2-117^2}{2*1500}=-3.42\:\rm ft/s^2$$