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A young gazelle is grazing in a beautifully sunlit corner of the savanna. Suddenly, the gazelle raises herhead and spots a lioness in the tall grass 173 m away, so she turns away running at roughly constantspeed. The lioness immediately chases the gazelle, with an explosive acceleration of 2.57 m/s2. Aninformed source tells us that this lioness is capable of enduring her maximum speed of 21.0 m/s for 25.0seconds at the longest. [Assume that both predator and prey never change their direction or motion inthis case, for the sake of simplicity.]
(a) On average, how fast (at least) does that gazelle need to run to survive? [Show all your steps.]
(b) Produce a qualitative graph of animal's position (vertical axis) versus time (horizontal axis), shared for the lioness's motion and as well as the gazelle's. Place graph labels in a way that fits the narration.

Answer :

Answer:

a)       v₂ = 13.20 m / s

Explanation:

To solve this exercise we will use the kinematic relations

Let's start with the Lioness.  Let's find the time to reach top speed

           v = v₀ + a t

as part of rest, its initial velocity is zero

           t = v / a

           t₁ = 21.0 / 2.57

           t₁ = 8.17 s

the total time is the acceleration time plus the time (t₂ = 25 s) that the maximum speed can withstand

          t = t₁ + t₂

          t = 8.17 +25.0 = 33.17 s

Now let's find out what distance the lioness travels in these times

       

during acceleration

         x = v₀ t + ½ a t²

         x = ½ a t²

         x₁ = ½ 2.57 8.17²

         x₁ = 85.77 m

during constant speed part

        x₂ = v t₂

        x₂ = 21.0 25.0

        x₂ = 525 m

therefore the total distance traveled is

        x = x₁ + x₂

        x = 85.77 + 525

        x = 610.77 m

a) the average speed of the gazelle

this must be the distance that the lioness travels minus the initial distance that separates the two animals (xo = 173 m) between the time taken

        v₂ = [tex]\frac{x -x_o}{t}[/tex]

        v₂ = [tex]\frac{610.77 - 173}{33.17}[/tex]

        v₂ = 13.20 m / s

b) in the attachment we can see a graph of the displacement of the two animals

${teks-lihat-gambar} moya1316

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