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 =

Explanation:

Step1: Identify the initial - position and velocity

The ball starts 6 feet from Brian (initial position) and has a horizontal velocity of 15 feet per second. The distance function $d(t)$ from Brian is $d(t)=\vert6 - 15t\vert$ since the ball is moving towards Brian (negative - direction of the distance from Brian).

Step2: Set up the equation for the desired distance

We want to find when the ball is 3 feet from Brian, so we set $\vert6 - 15t\vert=3$.

Step3: Solve the absolute - value equation

Case 1: $6−15t = 3$.
Subtract 6 from both sides: $-15t=3 - 6=-3$.
Divide both sides by - 15: $t=\frac{-3}{-15}=\frac{1}{5}=0.2$.
Case 2: $6−15t=-3$.
Subtract 6 from both sides: $-15t=-3 - 6=-9$.
Divide both sides by - 15: $t=\frac{-9}{-15}=\frac{3}{5}=0.6$.

Answer:

The ball is 3 feet horizontally from Brian at $t = 0.2$s and $t = 0.6$s.