A gazelle is running at a constant speed of 19.3 m/s toward a motionless hidden cheetah. At the instant the gazelle passes the cheetah, the cheetah accelerates at a rate of 7.1 m/s/s in pursuit of the gazelle. The gazelle maintains its constant speed. By the time the cheetah reaches a speed of 19.3 m/s to match the gazelle, how far apart are the two animals in units of m

the animals are 26.2 meters apart.

Explanation:

Let's define t = 0s as the moment when the cheetah starts accelerating.

The gazelle moves with constant velocity, thus, it is not accelerating, then the acceleration of the gazelle is:

a₁(t) = 0m/s^2

where I will use the subscript "1" to refer to the gazelle and "2" to refer to the cheetah.

for the velocity of the gazelle we just integrate over time to get:

v₁(t) = V0

where V0 is the initial speed of the gazelle, which we know is 19.3 m/s

v₁(t) = 19.3 m/s

To get the position of the gazelle we integrate again:

p₁(t) = ( 19.3 m/s)*t + P0

where P0 is the position of the gazelle at t = 0s, let's define P0 = 0m

p₁(t) = ( 19.3 m/s)*t

The equations that describe the motion of the gazelle are:

a₁(t) = 0m/s^2

v₁(t) = 19.3 m/s

p₁(t) = ( 19.3 m/s)*t

Now let's do the same for the cheetah.

We know that its acceleration is 7.1 m/s^2

then:

a₂(t) =  7.1 m/s^2

for the velocity of the cheetah we integrate:

v₂(t) = (7.1 m/s^2)*t + V0

where v0 is the initial velocity of the cheetah, which we know its zero.

v₂(t) = (7.1 m/s^2)*t

Finally, for the position equation we integrate again, and remember that we have defined the initial position for the gazelle as zero, then the same happens for the cheetah.

p₂(t) = (1/2)*(7.1 m/s^2)*t^2

The equations for the cheetah are:

a₂(t) =  7.1 m/s^2

v₂(t) = (7.1 m/s^2)*t

p₂(t) = (1/2)*(7.1 m/s^2)*t^2

Now, we want to find the distance between both animals when the speed of the cheetah is 19.3 m/s, then first we need to solve:

v₂(t) = (7.1 m/s^2)*t =  19.3 m/s

t = (19.3 m/s)/(7.1 m/s^2) = 2.72s

Now, to find the distance between the two animals, we just compute the difference between the position equations for t = 2.72s

Distance = p₁(2.72s)  -  p₂(2.72s)

               = ( 19.3 m/s)*2.72s -  (1/2)*(7.1 m/s^2)*(2.72s)^2

               = 26.2 m

So the animals are 26.2 meters apart.

c

explanation:

i believe the answer is c and this is only worth 10 points you might wanna change it

blue light is scattered in different directions by the tiny molecules of air. blue is scattered more than other colours because it has a shorter distance to travel which results in smaller waves. this is why we see a blue sky most of the time.