maya is driving a racecar. the table below gives the distance ( d(t) ) (in meters) she has driven at a few…

maya is driving a racecar. the table below gives the distance ( d(t) ) (in meters) she has driven at a few times ( t ) (in seconds) after she starts.\n\n(a) find the average rate of change for the distance driven from 0 seconds to 5 seconds.\n\n( square ) meters per second\n\n(b) find the average rate of change for the distance driven from 7 seconds to 9 seconds.\n\n( square ) meters per second

maya is driving a racecar. the table below gives the distance ( d(t) ) (in meters) she has driven at a few times ( t ) (in seconds) after she starts.\n\n(a) find the average rate of change for the distance driven from 0 seconds to 5 seconds.\n\n( square ) meters per second\n\n(b) find the average rate of change for the distance driven from 7 seconds to 9 seconds.\n\n( square ) meters per second

Answer

Explanation:

Step1: Recall the formula for average rate of change

The formula for the average rate of change of a function (y = f(x)) over the interval ([x_1,x_2]) is (\frac{f(x_2)-f(x_1)}{x_2 - x_1}).

Step2: Solve part (a)

For the interval from (t_1 = 0) to (t_2=5), (D(t_1)=0) and (D(t_2) = 151.5). Using the formula (\frac{D(t_2)-D(t_1)}{t_2 - t_1}=\frac{151.5 - 0}{5-0}=\frac{151.5}{5}=30.3)

Step3: Solve part (b)

For the interval from (t_1 = 7) to (t_2 = 9), (D(t_1)=205.1) and (D(t_2)=255.9) Using the formula (\frac{D(t_2)-D(t_1)}{t_2 - t_1}=\frac{255.9 - 205.1}{9 - 7}=\frac{50.8}{2}=25.4)

Answer:

a) (30.3) meters per second b) (25.4) meters per second