Samantha and Isaac are playing racquetball. Samantha hits the ball, sending it on a trajectory modeled by y=−3|x−4|+20

Question

, where y is the height reached by the ball, in feet, after x milliseconds. In a desperate attempt to keep the ball in the air, Isaac throws his racquet toward it at a trajectory modeled by y=1/3x+4
. When does his racquet hit the ball?(1 point)
Responses

after 3.5 milliseconds
after 3.5 milliseconds

after 8.4 milliseconds
after 8.4 milliseconds

after 1.5 milliseconds
after 1.5 milliseconds

after 6.8 milliseconds

Answer 1

To determine when Isaac's racquet hits the ball, we need to find the intersection point of the two trajectories. We set the equations equal to each other:

1. Samantha's trajectory: \[ y = -3|x - 4| + 20 \]

2. Isaac's trajectory: \[ y = \frac{1}{3}x + 4 \]

We set these two equations equal to find the intersection:

\[ -3|x - 4| + 20 = \frac{1}{3}x + 4 \]

Next, we can consider two cases based on the absolute value function.

Case 1: \( x - 4 \geq 0 \) (which implies \( x \geq 4 \))

In this case, the equation simplifies to: \[ -3(x - 4) + 20 = \frac{1}{3}x + 4 \] \[ -3x + 12 + 20 = \frac{1}{3}x + 4 \] \[ -3x + 32 = \frac{1}{3}x + 4 \]

Multiplying everything by 3 to eliminate the fraction: \[ -9x + 96 = x + 12 \] \[ -10x = -84 \] \[ x = 8.4 \]

Case 2: \( x - 4 < 0 \) (which implies \( x < 4 \))

In this case, the equation simplifies to: \[ -3(-x + 4) + 20 = \frac{1}{3}x + 4 \] \[ 3x - 12 + 20 = \frac{1}{3}x + 4 \] \[ 3x + 8 = \frac{1}{3}x + 4 \]

Again, multiplying everything by 3: \[ 9x + 24 = x + 12 \] \[ 8x = -12 \] \[ x = -1.5 \]

This solution doesn't make sense in the context of time since it cannot be negative.

The only valid solution is from Case 1: Isaac's racquet hits the ball after 8.4 milliseconds.