Question

charley throws the ball back to alice with a horizontal velocity of 15 feet per second. find the times when the ball is 3 feet horizontally from brian. write a function for the distance of the ball from brian, as viewed from overhead. notice that the ball now starts 6 feet to brians , and moves in a direction. d(t)=| - t| now find the times the ball is 3 feet horizontally from brian. |6 - 15t|= 6 - 15t= or 6 - 15t= - 15t= or - 15t= t= s or t= s written as decimals, the ball is 3 feet horizontally from brian at t= s and at t= s.

Explanation:

Step1: Set up the distance - time equation

The ball starts 6 feet from Brian and has a horizontal velocity of 15 feet per second. The distance function is $d(t)=|6 - 15t|$. We want to find when $d(t) = 3$, so we set up the equation $|6-15t| = 3$.

Step2: Solve the absolute - value equation

An absolute - value equation $|a|=b$ ($b\geq0$) is equivalent to $a = b$ or $a=-b$. So, $6 - 15t=3$ or $6 - 15t=-3$.

Case 1: Solve $6 - 15t = 3$

Subtract 6 from both sides: $-15t=3 - 6=-3$. Then divide both sides by - 15: $t=\frac{-3}{-15}=\frac{1}{5}=0.2$.

Case 2: Solve $6 - 15t=-3$

Subtract 6 from both sides: $-15t=-3 - 6=-9$. Then divide both sides by - 15: $t=\frac{-9}{-15}=\frac{3}{5}=0.6$.

Answer:

$t = 0.2$ s and $t = 0.6$ s