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Famed stunt pilot, Cleonvia Thread, pulls out rapidly from a dive. He is traveling at 222 mi/h at the bottom of his trajectory, and at that instant is traveling on a curve of radius 820 ft. r v What acceleration does he

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asuwafohjames

Answer:

0.39 m/s²

Explanation:

From the question,

a = v²/r.................... Equation 1

Where v = velocity, r = radius.

Given: v = 222 mi/h = (222×0.44704) m/s = 9.83 m/s, r = 820 ft = (820×0.3048) m = 249.94 m.

Substitute thses values into equation 1

a = 9.83²/249.94

a = 96.63/249.94

a = 0.39 m/s²

Hence the  acceleration is 0.39 m/s²

The value of the centripetal acceleration is equal to 4g.

Data Given;

  • v = 222mi/h
  • distance (r) = 820ft

Centripetal Acceleration

Using formula of centripetal acceleration,

[tex]a_c = \frac{v^2}{r}\\ [/tex]

v = (222 * 1.467) ft/s

v = 325.6 ft/s

substitute the values into the formula

[tex]a_c = v^2/r\\ a_c = \frac{325.6^2}{820}\\ a_c = 129.28 ft/s^2\\ [/tex]

But the value of g = 32.174 ft/s62

This makes a = 4g.

Learn more on centripetal acceleration here;

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