Question

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) =

Answer 1

Answer:

a. Alice

b. Briana

c. 0.51 minutes

Step-by-step explanation:

a. Alice formula is valid for any t > 0 minutes, but Briana formula is only valid for

2t - 1 > 0

2t > 1

t > 1/2 minutes

b. They finish when their covered distance is equal to 5 kilometers. For Alice:

t/4 = 5

t = 5*4 = 20 minutes

For Briana:

√(2t - 1) = 5

2t - 1 = 5²

2t = 25 + 1

t = 26/2

t = 13 minutes

c. They are side by side when they have covered the same distance, that is:

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 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 answer has sense for this problem because the race already finished before they spent 31.49 minutes in it.

Answer 2

Answer:

a) Alice starts first, b) Briana gets the finish line first, c) $t \approx 0.509\,min$

Step-by-step explanation:

a) Alice starts first, since binomial inside the squared root is equal to zero when $t = \frac{1}{2}\,min$, whereas Alice begins at $t = 0\,min$.

b) The instant when each competitor reach 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 gets the finish line first.

c) Alice and Briana are side by side when $a(t) = b(t)$. Then:

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

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

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

The roots of the second order polynomial are:

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

Just the second roots makes sense, as they must be side by side at one instant before getting the finish line.