Question

jon and brianna were racing remote - control cars. the equation d = 3t+15 models the distance, d (in feet), jons car was away from the wall, after t seconds. the table below shows the distance that briannas car was away from the same wall at different rates.

seconds (t)	5	10	15

in feet, how much closer to the wall did briannas car start compared to jons car?

Explanation:

Step1: Find initial distance of Jon's car

When \(t = 0\), using the equation \(d=3t + 15\), substitute \(t = 0\). We get \(d=3\times0+15=15\) feet.

Step2: Find initial distance of Brianna's car

From the table, when \(t = 0\), the distance of Brianna's car from the wall \(d = 20\) feet.

Step3: Calculate the difference

To find how much closer Jon's car is to the wall initially, subtract the initial - distance of Jon's car from the initial - distance of Brianna's car. \(20−15 = 5\) feet.

Answer:

5 feet