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SCORPION-xisa
2 months 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) =

Mathematics
2 answers:
zzz [12.3K]2 months 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 [12.4K]2 months 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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