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5 days ago

12

Alice and Briana each participate in a 5 kilometer race. Alice's distance covered, in kilometers, after t minutes can be modeled

by the equation a(t) =

2 answers:

zzz [4K]5 days ago

4 0

Answer:

a) Alice moves first, b) Briana finishes the race ahead, c) $t \approx 0.509\,min$

Step-by-step explanation:

a) Alice gets underway first, since the binomial under the square root becomes zero when $t = \frac{1}{2}\,min$ , while Alice starts at $t = 0\,min$ .

b) The moment each runner crosses the finish line is:

Alice

$5\,km = \frac{t}{4}$

$t = 20\,min$

Briana

$5\,km = \sqrt{2\cdot t - 1}$

$25\,km^{2} = 2\cdot t - 1$

$t = 13\,min$

Briana reaches the finish before Alice.

c) Alice and Briana are at the same point when $a(t) = b(t)$ . From here:

$\frac{t}{4} = \sqrt{2\cdot t - 1}$

$\frac{t^{2}}{16} = 2\cdot t - 1$

$t^{2} - 32\cdot t + 16 = 0$

The roots of the polynomial are:

$t_{1} \approx 31.492\,min$ and $t_{2} \approx 0.509\,min$

Only the second root is relevant, as they must be parallel at some point prior to finishing the race.

PIT_PIT [3.9K]5 days ago

4 0

Answer:

a. Alice

b. Briana

c. 0.51 minutes

Step-by-step explanation:

a. Alice's formula is applicable for any t > 0 minutes, while Briana’s is only applicable when

2t - 1 > 0

2t > 1

t > 1/2 minutes

b. They complete the race when their total distance equals 5 kilometers. For Alice:

t/4 = 5

t = 20 minutes

For Briana:

√(2t - 1) = 5

2t - 1 = 25

2t = 26

t = 13 minutes

c. They reach the same point when they have traveled the same distance, expressed as:

t/4 = √(2t - 1)

(t/4)² = 2t - 1

t²/16 = 2t - 1

t² = 16(2t - 1)

t² = 32t - 16

t² - 32t + 16 = 0

Using the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4(a)(c)}}{2(a)}$

$t = \frac{32 \pm \sqrt{-32^2 - 4(1)(16)}}{2(1)}$

$t = \frac{32 \pm 30.98}{2}$

$t_1 = \frac{32 + 30.98}{2}$

$t_1 = 31.49$

$t_2 = \frac{32 - 30.98}{2}$

$t_2 = 0.51$

Only the second solution fits this scenario because the race concluded before they took 31.49 minutes.

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