Question

The average speed of a car decreased by 3 miles per hour every successive 8-minute interval. If the car traveled 4.8 miles in the sixth 8-minute interval, what was the average speed of the car, in miles per hour, in the first 8-minute interval?

Answer 1

Answer:

13.5 mi/h

Step-by-step explanation:

The average speed of the car can be written as

$v=\frac{d}{t}$

where

d = 4.8 miles is the total distance covered

$t=6\cdot (8 min) = 48 min \cdot \frac{1}{60}=0.8 h$ is the  time elapsed

So the average speed is

$d=\frac{4.8}{0.8}=6 mi/h$

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 $v_1$ the average speed in the 1st interval, we have:

$v_2=v_1-3\\v_3=v_1-6\\v_4=v_1-9\\v_5=v_1-12\\v_6=v_1-15$

The average speed in each interval can be written as $v_i=\frac{d_i}{t}$, where $d_i$ is the distance covered in each interval and $d_i = 8 min =0.133 h$ is the duration of each interval, so we can write

$\frac{d_2}{t}=\frac{d_1}{t} -3 \rightarrow d_2 = d_1 -3t$

And similarly,

$d_3=d_1-6t\\d_4=d_1-9t\\d_5=d_1-12t\\d_6=d_1-15t$

Since the total distance is $d=d_1+d_2+d_3+d_4+d_5+d_6$, we have:

$d=6d_1 - 3t-6t-9t-12t-15t=6d_1-45t$

And since we know that

d = 4.8 miles

and t = 0.133h, we can find d1:

$d_1=\frac{d+45t}{6}=\frac{4.8+(45)(0.133)}{6}=1.8 mi$

So the average speed in the first 8-minute interval is:

$v_1=\frac{d_1}{t_1}=\frac{1.8}{0.133}=13.5 mi/h$