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2 years ago

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Cheetahs, the fastest of the great cats, can reach 45 mph in 2.0 sec starting from rest. Assuming that they have constant accele

ration throughout that time, find (a) their acceleration (in f t/s2 and m/s2 ) (b) the distance they travel during that time (in m and ft)

1 answer:

Kipish [7]2 years ago

7 0

Answer:

acceleration is 10.05 m/s² or 32.97 ft/s²

distance is 40.22 m or 131.95 ft

Explanation:

given data

velocity = 45 mph = 20.1168 m/s

time = 2 sec

to find out

acceleration and distance

solution

first we will apply here formula for acceleration that is

v = u +at ........................1

here v is velocity and u is initial speed and t is time and a is acceleration

so put here all value

20.11 = 0 + a(2)

a = 10.05 m/s

so acceleration is 10.05 m/s² or 32.97 ft/s²

and

distance is calculated as

distance = v × t ..................2

put here value

distance = 20.11 × 2

distance = 40.22

so distance is 40.22 m or 131.95 ft

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Details are missing in the question. Complete text of the problem:

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$v=\omega A$

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For this system, we have

$\omega=2\pi f=2\pi (0.84 Hz)=5.25 rad/s$

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Substituting into the formula, we find the maximum speed:

$v=(5.25 rad/s)(0.200 m)=1.05 m/s$

(d) x = 0

The maximum speed in a simple harmonic motion occurs at the equilibrium position. In fact, the total mechanical energy of the system is equal to the sum of the elastic potential energy (U) and the kinetic energy (K):

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Using the numbers we calculated in part c):

$\omega=2\pi f=2\pi (0.84 Hz)=5.25 rad/s$

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we find immediately the maximum acceleration:

$a=(5.25 rad/s)^2(0.200 m)=5.51 m/s^2$

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this means that the acceleration will be maximum when the force is maximum.

However, the force is given by Hooke's law:

$F=kx$

so, the force is maximum when the displacement x is maximum: so, the maximum acceleration occurs at the position of maximum displacement.

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$E=K=\frac{1}{2}mv_{max}^2$

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m = 2.90 kg is the mass

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Solving for E, we find

$E=\frac{1}{2}(2.90 kg)(1.05 m/s)^2=1.60 J$

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$x=\frac{1}{3}(0.200 m)=0.0667 m$

so the elastic potential energy is

$U=\frac{1}{2}kx^2=\frac{1}{2}(80 N/m)(0.0667 m)^2=0.18 J$

and since the total energy E = 1.60 J is conserved, the kinetic energy is

$K=E-U=1.60 J-0.18 J=1.42 J$

And from the relationship between kinetic energy and speed, we can find the speed of the system:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1.42 J)}{2.90 kg}}=0.99 m/s$

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$x=\frac{1}{3}(0.200 m)=0.0667 m$

So the restoring force exerted by the spring on the mass is

$F=kx=(80 N/m)(0.0667 m)=5.34 N$

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$a=\frac{F}{m}=\frac{5.34 N}{2.90 kg}=1.84 m/s^2$

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