To solve this, you have to find the interval in which d > 0. For starters, solve for the x-intercepts of 0, or when d=0. [tex]d=144-16t^2=\ \textgreater \ 144-d=16t^2=\ \textgreater \ \frac{144-d}{16} =t^2=\ \textgreater \ |t|= \sqrt{\frac{144-d}{16}} [/tex]. You can now plug in d=0. [tex]|t|= \sqrt{\frac{144-0}{16}} = \sqrt{\frac{144}{16}}=\sqrt{9}=3[/tex]. Now, as |x|=y is the same as x=±y, t=±3 or t=3 and t=-3. Next, we determine if our d is positive or negative in the interval (-∞,∞) with subintervals at our x-intercepts, making our intervals (-∞,-3), (-3,3), and (3,∞). To do this, just take one value from each interval and plug it in for t. For interval (-∞,-3), we can use -4 so [tex]d=144-16(-4)^2=144-16(16)=144-256=-112[/tex], making all ds in this interval negative. For (-3,3), the easiest thing to use is 0 so [tex]d=144-16(0)^2=144[/tex], making all ds in this interval positive. For interval (3,∞), we can use 4 so [tex]d=144-16(4)^2=144-16(16)=144-256=-112[/tex], making all ds in this interval negative. As we need d>0, we can conclude that in the interval (-3,3) the car is in the air. Lastly, we must consider that t cannot be less than 0 as there is no such thing as negative time, so with 0 as our domain restriction, we can conclude the interval in which the car is in the air is (0,3), also written as t ∈ (0,3).
