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.