"At a cross-country track meet, Alicia ran 8 mph for the first part of the race, then increased her speed to 12 mph for the second part. If the race was 21 miles long and Alicia finished in 2 hours, how far did she run at the faster pace?"

Respuesta :

Answer:

15 miles

Explanation:

Alicia's speed in the first part of the race is:

[tex]v_1 = \frac{d_1}{t_1}=8 mph[/tex]

where [tex]d_1[/tex] is the length of the first part and [tex]t_1[/tex] the time taken to cover this part.

Alicia's speed in the second part of the race is:

[tex]v_2 = \frac{d_2}{t_2}=12 mph[/tex]

where [tex]d_2[/tex] is the length of the second part and [tex]t_2[/tex] the time taken to cover this part.

The total length of the race is

[tex]d=d_1 +d_2 = 21 mph[/tex]

And the total time taken by Alicia to complete the race is

[tex]t=t_1 + t_2 = 2 h[/tex]

We have in total 4 equations with 4 unknown quantities. From the first 2 equations we find:

[tex]d_1 = 8t_1\\d_2 = 12 t_2[/tex]

And substituting into the second group of equations:

[tex]8t_1 + 12t_2 =21\\t_1 + t_2 = 2[/tex]

Calling A the first equation and B the second equation, if we compute (A-8B) we get:

[tex]4t_2 = 5 \rightarrow t_2 = 1.25h[/tex]

And so

[tex]t_1 = 2-t_2 = 0.75 h[/tex]

So Alicia runs for 1.25 hours at 12 mph (second part of the race). Therefore, the distance travelled during this part is:

[tex]d_2 = v_2 t_2 = 12 \cdot 1.25 = 15 mi[/tex]

Therefore the answer is 15 miles.

Alicia ran 15 miles at the faster pace.

Formulating the equations:

Let Alicia run with the speed of 12mph for A hours and with the speed of 8mph for B hours.

Total distance is 21 miles, so we can make the first equation as:

12A + 8B = 21      .............(i)

Now the total time taken is 2 hours, which means:

A + B = 2             ..............(ii)

Now multiply equation (ii) by 8 and subtract from equation (i)

then we get,   4A = 5

A = 5/4 hours

So Alicia run at the speed of 12mph for 5/4 hours, total distance covered is:

d = 12 × 5/4

d = 15 miles

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