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faltersainse [42]
3 years ago
5

Find the maximum walking speed of giraffe with leges 6.03 feet long

Mathematics
1 answer:
Rasek [7]3 years ago
6 0

Answer:

13.89 ft/s

Step-by-step explanation:

We know that the cheetah is the fastest land animal on Earth, but what if it were not allowed to run? Biologists model maximum walking speed by the equation = √, where g is the acceleration due to gravity (9.8 m/s2 or 32 ft/s2) and L is the length of the leg (in metric or standard units). In the situations below, use this information to answer the questions.  

(Hint: It will be useful to know 1 m = 3.23 ft for conversions)  

 

1. Find the maximum walking speed of a giraffe with legs 6.03 feet long.  

Solution:

Speed is the ratio of the total distance travelled to the time taken to reach this distance. Speed is given by the formula:

Speed = distance / time

Given that the maximum walking speed (s) is given by the equation:

= √, where g is the acceleration due to gravity (9.8 m/s² or 32 ft/s²) and L is the length of the leg (in metric or standard units).

The giraffe has a length of leg (L) of 6.03 feet, hence:

s = √(32 ft/s² * 6.03 ft)

s = √192.96

s = 13.89 ft/s

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Step-by-step explanation:

Drawing the right triangle (as attached) gives us that \arctan \left(\frac{12}{5} \right)=\arcsin \left(\frac{12}{13} \right)

Also, -\arcsin \left(-\frac{3}{5} \right)=\arcsin \left(\frac{3}{5} \right)

This means our original expression is equal to:

\cos \left[\arcsin \left(\frac{12}{13} \right)+\arcsin \left(\frac{3}{5} \right) \right]

Using the cosine addition formula, which states \cos(a+b)=\cos a \cos b-\sin a \sin b, we get this itself is equal to:

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Since \sin^{2} \theta+\cos^{2} \theta=1, we know that:

\sin^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)+\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=1\\\\\frac{144}{169} +\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=1\\\\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=\frac{25}{169}\\\\cos \left(\arcsin \left(\frac{12}{13} \right)\right)=\frac{5}{13}

Similarly, cos(arcsin(3/5))=4/5.

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\left(\frac{5}{13} \right) \left(\frac{4}{5} \right)-\left(\frac{12}{13} \right) \left(\frac{3}{5} \right)\\\\\frac{20}{65}-\frac{36}{65}=\boxed{-\frac{16}{65}}

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