The world's fastest land animal, the cheetah, can accelerate at 8.15 m/s2 to a top speed of vcmax = 33 m/s. a cheetah observes a passing gazelle traveling at vg = 18.4 m/s and begins to chase it. how long, in seconds, does it take the cheetah to reach its maximum velocity, vcmax, assuming its acceleration is constant?

t=4.04s

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t  (1)

{Vf^{2}-Vo^2}/{2.a} =X(2)

X=Xo+ VoT+0.5at^{2}    (3)\\

X=(Vf+Vo)T/2 (4)

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve

to solve this problem we can use the ecuation number 1

Vf=33m/s

Vo=0m/s, ( the cheetah starts from rest, although the gazelle is moving=

a=8.15m/s^2)

Vf=Vo+at

t=(Vf-Vo)/a

t=(33-0)/8.15

t=4.04s

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