*1 A car is driving away from a crosswalk. The distance d (in feet) of the car from the crosswalk tseconds since the car started moving is given by the formula d = t^2+3.5.a. As the number of seconds since the car started movingincreases from 1 second to 3 seconds, what is the changein the car's distance from the crosswalk?
Accepted Solution
A:
Answer:The change in the car's distance is 8 feet Step-by-step explanation:* Lets explain how to solve the problem- A car is driving away from a crosswalk- The distance d (in feet) of the car from the crosswalk t seconds since the car started moving is given by the formula d = t² + 3.5- The time increasing from 1 second to 3 seconds- We need to now the change of the car's distance from the crosswalk∵ The equation of the distance is d = t² + 3.5∵ The time is 1 second∴ d = (1)² + 3.5∴ d = 1 + 3.5 = 4.5 feet∵ The time is 3 seconds∴ d = (3)² + 3.5∴ d = 9 + 3.5 = 12.5 feet∵ The change of the distance = d of 3 sec - d of 1 sec∵ d of 3 sec = 12.5 feet∵ d of 1 sec = 4.5 feet∴ The change of the distance = 12.5 - 4.5 = 8 feet∴ The change in the car's distance is 8 feet