Question

Starting from 300 feet away, a car drives toward you. It then passes by you at a speed of 48 feet per second. The distance d (in feet) of the car from you after t seconds is given by the equation ​d=|300−48t|. At what times is the car 60 feet from you? needs two answers

Answer 1

$Solution,60=\left|300-48t\right|\quad :\quad t=5\quad \mathrm{or}\quad \:t=\frac{15}{2}$

$Steps:$

$60=\left|300-48t\right|$

$\mathrm{Switch\:sides},\\\left|300-48t\right|=60,\\\left|300-48t\right|=60$

$|f\left(t\right)|=a\quad \Rightarrow \:f\left(t\right)=-a\quad \mathrm{or}\quad \:f\left(t\right)=a,\\300-48t=-60\quad \quad \mathrm{or}\quad \:\quad \:300-48t=60$

$300-48t=-60,\\\mathrm{Subtract\:}300\mathrm{\:from\:both\:sides},\\300-48t-300=-60-300,\\\mathrm{Simplify},\\-48t=-360,\\\mathrm{Divide\:both\:sides\:by\:}-48,\\\frac{-48t}{-48}=\frac{-360}{-48},\\\mathrm{Simplify},\\t=\frac{15}{2}$

$300-48t=60,\\\mathrm{Subtract\:}300\mathrm{\:from\:both\:sides},\\300-48t-300=60-300, \\\mathrm{Simplify}, \\-48t=-240, \\\mathrm{Divide\:both\:sides\:by\:}-48,\\\frac{-48t}{-48}=\frac{-240}{-48}, \\\mathrm{Simplify}, \\t=5$

$\mathrm{Combine\:the\:ranges},\\t=5\quad \mathrm{or}\quad \:t=\frac{15}{2}$

$\mathrm{The\:Correct\:Answer\:is\:t=5\quad \mathrm{or}\quad \:t=\frac{15}{2}}$

$\mathrm{Hope\:This\:Helps!!!}$

$\mathrm{-Austint1414}$