Answer:
13.5 mi/h
Step-by-step explanation:
The average speed of the car can be written as
[tex]v=\frac{d}{t}[/tex]
where
d = 4.8 miles is the total distance covered
[tex]t=6\cdot (8 min) = 48 min \cdot \frac{1}{60}=0.8 h[/tex] is the time elapsed
So the average speed is
[tex]d=\frac{4.8}{0.8}=6 mi/h[/tex]
We also know that the total time consists of 6 8-minutes interval, and the speed of the car decreased by 3 mi/h each interval.
Calling [tex]v_1[/tex] the average speed in the 1st interval, we have:
[tex]v_2=v_1-3\\v_3=v_1-6\\v_4=v_1-9\\v_5=v_1-12\\v_6=v_1-15[/tex]
The average speed in each interval can be written as [tex]v_i=\frac{d_i}{t}[/tex], where [tex]d_i[/tex] is the distance covered in each interval and [tex]d_i = 8 min =0.133 h[/tex] is the duration of each interval, so we can write
[tex]\frac{d_2}{t}=\frac{d_1}{t} -3 \rightarrow d_2 = d_1 -3t[/tex]
And similarly,
[tex]d_3=d_1-6t\\d_4=d_1-9t\\d_5=d_1-12t\\d_6=d_1-15t[/tex]
Since the total distance is [tex]d=d_1+d_2+d_3+d_4+d_5+d_6[/tex], we have:
[tex]d=6d_1 - 3t-6t-9t-12t-15t=6d_1-45t[/tex]
And since we know that
d = 4.8 miles
and t = 0.133h, we can find d1:
[tex]d_1=\frac{d+45t}{6}=\frac{4.8+(45)(0.133)}{6}=1.8 mi[/tex]
So the average speed in the first 8-minute interval is:
[tex]v_1=\frac{d_1}{t_1}=\frac{1.8}{0.133}=13.5 mi/h[/tex]
