Question

Jon and Brianna were racing remote control cars. The equation d-3t+

15 models the distance, d (in feet), Jon's car was away from a wall, after
t seconds. The table below shows the distance that Brianna's car was
away from the same wall at different times.
Feet from Wall (d)
20
30
40
How much closer to the wall did Brianna's car start compared to Jon's car?
A
8
C
1 ft
2 ft
5ft
10 ft
Seconds (t)
5
10
15

Answer 1

Brianna's car started 5 feet closer to the wall compared to Jon's car.

Step-by-step explanation:

To determine how much closer to the wall Brianna's car started compared to Jon's car, we need to find the initial distance for each car.

For Jon's car, the initial distance can be found by plugging in t=0 into the equation d-3t+15.

This gives us d-3(0)+15 = d+15. So Jon's car started at a distance of d+15 feet from the wall.

For Brianna's car, we can refer to the table where it shows the distance from the wall at different times. The first value in the table, 20, represents the distance at t=0.

Therefore, Brianna's car started at a distance of 20 feet from the wall.

To find how much closer Brianna's car started compared to Jon's car, we subtract Brianna's initial distance from Jon's initial distance: (d+15) - 20 = d-5.

Therefore, Brianna's car started 5 feet closer to the wall compared to Jon's car.