How Much Power Does a Cheetah Generate During Acceleration?

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Homework Help Overview

The problem involves calculating the average power generated by a cheetah during its acceleration from rest to a speed of 3.4 m/s over a time span of 0.5 seconds, with a given mass of 112 kg. The context is rooted in physics, specifically in dynamics and power calculations.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various methods to calculate power, including using kinematic equations and the relationship between work and energy. Some raise questions about unit consistency and the validity of their calculations.

Discussion Status

There is an ongoing exploration of different approaches to determine power, with some participants suggesting alternative methods and questioning the assumptions made in the calculations. A few participants express uncertainty about their results, while others acknowledge corrections made during the discussion.

Contextual Notes

Some participants question the acceleration rate of the cheetah and compare it to their own potential performance, indicating a broader discussion about the implications of the cheetah's acceleration capabilities.

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Homework Statement



The cheetah is one of the fastest-accelerating animals, because it can go from rest to 3.4 m/s (about 8 mi/h) in 0.5 s. If its mass is 112 kg, determine the average power developed by the cheetah during the acceleration phase of its motion. Express your answer in watts and horsepower.


Homework Equations





The Attempt at a Solution



v = u + at ...since the cheetah is initially at rest , u = 0 m/s therefore
v = at and we have a = (v/t)

Now power is defined as rate of doing work

P = ( Force x displacement ) / time
P = (ma x s) / t where s is the displacement s = vt
P = (m) * (v/t) (vt) divided by time t

finally doing all the algebra gives us

P = mv ^2/ t
P = (112 x 3.4^2) / 0.5
P = 2589.4

are my units wrong? but where?
 
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Displacement s = average velocity* time.
So s = vt/2
 
Or you could simply find the change in KE and power = change in ke/time
 
so would it be p= {m(v/t)(vt/2)}/t
p= 1294.7 watts

is that right?
 
triplel777 said:
so would it be p= {m(v/t)(vt/2)}/t
p= 1294.7 watts

is that right?
No. It is only m*v^2.
Power = 1/2*m*v^2/t
 
rl.bhat said:
No. It is only m*v^2.
Power = 1/2*m*v^2/t

ok so 1/2*m*v^2/t
=1/2*112*3.4^2/0.5
=1294.7

is that it?
 
Oh! Sorry. You are right.
 
yup thanks so much :)
 
triplel777 said:

Homework Statement



The cheetah is one of the fastest-accelerating animals, because it can go from rest to 3.4 m/s (about 8 mi/h) in 0.5 s.

Really? I'm pretty sure if I started with a big leap, I could hit 8mph in not much more than .5s. Is that a high acceleration?

"...a cheetah can go from 0 to 60 miles (96 kilometers) an hour in only three seconds..."
http://animals.nationalgeographic.com/animals/mammals/cheetah.html

So, 0 to 8 in the first .5s, then 8 to 60 in 2.5s.

OK, that's nearly the same slope... (8/.5 = 16) versus (52/2.5 = 20.8).
 
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