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.

no. dot product   =0 when vecs at right angles

the answer is a. hope this .