A speedy rabbit is hopping to the right with a velocity of 4.0 \,\dfrac{\text m}{\text s}4.0 s m 4, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction when it sees a carrot in the distance. The rabbit speeds up to its maximum velocity of 13 \,\dfrac{\text m}{\text s}13 s m 13, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction with a constant acceleration of 2.0 \,\dfrac{\text m}{\text s^2}2.0 s 2 m 2, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction rightward.

In this situation we need to find the distance [tex]d[/tex] between the rabbit and the carrot, and we can use the following equation, since the rabbit's acceleration is constant:

[tex]V^{2}=V_{o}^{2} + 2ad[/tex] (1)

Where:

[tex]V=13 m/s[/tex] is the rabbit's maximum velocity (final velocity)

[tex]V_{o}=4 m/s[/tex] is the rabbit's initial velocity

[tex]a=2 m/s^{2}[/tex] is the rabbit's acceleration

[tex]d[/tex] is the distance between the rabbit and the carrot

Isolating [tex]d[/tex]:

[tex]d=\frac{V^{2}-V_{o}^{2}}{2a}[/tex] (2)

[tex]d=\frac{(13 m/s)^{2}-(4 m/s)^{2}}{2(2 m/s^{2})}[/tex] (3)

Finally:

[tex]d=38.25 m[/tex]

silvwerbubbles silvwerbubbles · Answer 2 · 2020-05-06T18:59:52+02:00

silvwerbubbles silvwerbubbles

06-05-2020

Answer:

4.5s

Explanation:

Cause that's what it says on my test hints

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