Question

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.

Answer 1

Answer: 38.25 m

Explanation:

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

$V^{2}=V_{o}^{2} + 2ad$ (1)

Where:

$V=13 m/s$ is the rabbit's maximum velocity (final velocity)

$V_{o}=4 m/s$ is the rabbit's initial velocity

$a=2 m/s^{2}$ is the rabbit's acceleration

$d$ is the distance between the rabbit and the carrot

Isolating $d$:

$d=\frac{V^{2}-V_{o}^{2}}{2a}$ (2)

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

Finally:

$d=38.25 m$

Answer 2

Answer:

4.5s

Explanation:

Cause that's what it says on my test hints