A cheetah looks up and sees a gazelle in the distance. The cheetah starts running towards the gazelle; when it catches the gazelle, it is running at the speed of 0.029km/s. If the acceleration of the cheetah is 4.14m/s2, how long does it take for the cheetah to catch the gazelle?

Get a new teacher, who does not use indefinite language: who is "it" in each instance? The world wonders.

I'm assuming the cheetah is running at 0.029 km/s (29m/s).

V = Vo + a*t = 29.
0 + 4.14*t = 29. t = ?.

To find the time it takes for the cheetah to catch the gazelle, we can use the equation:

v = u + at

Where:
v = final velocity of the cheetah (0.029 km/s)
u = initial velocity of the cheetah (0 m/s, assuming it starts from rest)
a = acceleration of the cheetah (4.14 m/s^2)
t = time taken to catch the gazelle (unknown)

First, we need to convert the final velocity from km/s to m/s:
0.029 km/s x 1000 m/km = 29 m/s

Now, we can substitute the values into the equation:

29 m/s = 0 m/s + (4.14 m/s^2) * t

Rearranging the equation to solve for t:

t = (29 m/s - 0 m/s) / (4.14 m/s^2)
t = 29 m/s / 4.14 m/s^2

Calculating the value of t:

t = 7 seconds

Therefore, it takes the cheetah 7 seconds to catch the gazelle.

To find the time it takes for the cheetah to catch the gazelle, we can use the equation of motion:

vf = vi + at

Where:
- vf is the final velocity (0.029 km/s)
- vi is the initial velocity (0 m/s, since the cheetah starts from rest)
- a is the acceleration (4.14 m/s^2)
- t is the time it takes (unknown)

First, let's convert the final velocity from km/s to m/s:

0.029 km/s * (1000 m/km) = 29 m/s

Now we can plug the values into the equation and solve for t:

29 m/s = 0 m/s + (4.14 m/s^2) * t

29 m/s = 4.14 m/s^2 * t

Now, rearrange the equation to solve for t:

t = 29 m/s / 4.14 m/s^2

t ≈ 7.014 seconds

Therefore, it takes approximately 7.014 seconds for the cheetah to catch the gazelle.