Answer :
a. A model for the difference in feet between car A and car B is d(t) = = 0.35t² - 3t - 1.
b. The difference in feet between car A and car B after 10 seconds is 4 feet
c. To tell which car is in the lead,
- If d(t) > 0, car A is in the lead but
- if d(t) < 0 car B is in the lead.
a. Write a model for the difference in feet between car A and car B.
Since the distance in feet of car A from the starting line is modeled by the function f(t) = 0.25t² + 22t - 1, where t is the time in seconds since the start of the race.
Also, the distance in feet of car B from the starting line is modeled by the function g(t) = -0.1t² + 25t.
The difference in distance between car A and car B d(t) is the difference between their distance from the starting line.
So, d(t) = f(t) - g(t)
= 0.25t² + 22t - 1 - ( -0.1t² + 25t)
= 0.25t² + 22t - 1 + 0.1t² - 25t
= 0.35t² - 3t - 1
So, a model for the difference in feet between car A and car B is d(t) = = 0.35t² - 3t - 1.
b. What is the difference in feet between car A and car B after 10 seconds?
Since the difference in distance in feet between car A and car B is d(t) = 0.35t² - 3t - 1.
To find the difference in distance after 10 seconds, we substitute t = 10 into d(t).
So, d(t) = 0.35t² - 3t - 1.
d(10) = 0.35(10)² - 3(10) - 1
= 0.35(100) - 30 - 1
= 35 - 31
= 4 feet
So, the difference in feet between car A and car B after 10 seconds is 4 feet
c. How can you tell which car is in the lead?
Since the difference in distance in feet between car A and car B is d(t) = 0.35t² - 3t - 1
= f(t) - g(t)
Since
- f(t) = distance of car A and
- g(t) = distance of car B.
So, to tell which car is in the lead,
- If d(t) > 0, car A is in the lead but
- if d(t) < 0 car B is in the lead.
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