Answer:
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.
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