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

4.5s

Step-by-step explanation:

Cause that's what it says on my test hints

Answer 2

Answer: 38.25 m

Step-by-step 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=(V^(2)-V_(o)^(2))/(2a)` (2)

`d=((13 m/s)^(2)-(4 m/s)^(2))/(2(2 m/s^(2)))` (3)

Finally:

`d=38.25 m`