Cheetah Acceleration: Solving Homework Problem in 3.2s

[12345ME]

A cheetah is hunting. Its prey runs for 3.2 s at a constant velocity of +10.0 m/s. Starting from rest, what constant acceleration must the cheetah maintain in order to run the same distance as its prey runs in the same time?

 

[kuruman]

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[tomcue]

I would like to clarify that the term "constant acceleration" in this context refers to a change in velocity that occurs at a constant rate over time. In other words, the cheetah needs to maintain a constant increase in its speed over the 3.2 seconds, rather than maintaining a constant speed throughout.

To solve this problem, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity (in this case, 0 m/s), a is the acceleration, and t is the time. We know that the prey's final velocity is +10.0 m/s and the time is 3.2 seconds. So, plugging in these values we get:

10.0 m/s = 0 m/s + a(3.2 s)

Solving for a, we get a = 3.125 m/s^2. This means that the cheetah needs to maintain a constant acceleration of 3.125 m/s^2 in order to run the same distance as its prey in the same time.

It is important to note that this calculation assumes that the cheetah is able to maintain a constant acceleration over the entire 3.2 seconds, which may not be the case in real life. Other factors such as terrain, obstacles, and the cheetah's own physical limitations may affect its acceleration. Nevertheless, this calculation provides a theoretical answer to the problem presented.