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.
How many seconds does it take the rabbit to speed up from 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 to 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?

Question

contestada

Respuesta :

skyluke89 skyluke89 · Answer 1 · 2020-04-10T09:50:52+02:00

Answer:

4.5 s

Explanation:

The motion of the rabbit is a uniformly accelerated motion (=at constant acceleration), therefore we can use the following suvat equation:

[tex]v=u+at[/tex]

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time elapsed

In this problem:

u = 4.0 m/s is the initial velocity of the rabbit

v = 13.0 m/s is the final velocity of the rabbit

[tex]a=2.0 m/s^2[/tex] is the acceleration of the rabbit

Solving for t, we find how many seconds does the rabbit take to accelerate to 13 m/s:

[tex]t=\frac{v-u}{a}=\frac{13-4}{2}=4.5 s[/tex]