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dlinn [17]
3 years ago
9

A world record was set for the men’s 100-m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt "coasted" ac

ross the finish line with a time of 9.69 s. If we assume that Bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, calculate his maximum speed and his acceleration. (b) During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race?
Mathematics
1 answer:
Andrei [34K]3 years ago
8 0

Answer:

a) v = 12.21m/s

a = 4.07 m/s²

b)v = 11.24m/s

a =  3.75 m/s²

Step-by-step explanation:

a) Dividing the moviment into two parts:

I - With acceleration

v = v₀ + at

s = s₀ + v₀t + at²/2

  • v₀ = 0
  • s₀ = 0
  • a = ?
  • v = ?
  • t = 3s
  • s = x

v = v₀ + at

v = 3a ⇒ a = v/3

s = s₀ + v₀t + at²/2

x = v/3.3²/2

x = 3v/2

II - Uniform

s = s₀ + vt

s = 100

s₀ = x

v = v

t = 9.69 - 3 = 6.69s

s = s₀ + vt

100 = x + v*6.69

100 = x + 6.69v

As x = 3v/2

100 = 3v/2 + 6.69v

100 = 1.5v + 6.69v

100 = 8.19v

v = 12.21m/s

a = v/3 = 4.07 m/s²

b) Dividing the moviment into two parts:

I - With acceleration

v = v₀ + at

s = s₀ + v₀t + at²/2

  • v₀ = 0
  • s₀ = 0
  • a = ?
  • v = ?
  • t = 3s
  • s = x

v = v₀ + at

v = 3a ⇒ a = v/3

s = s₀ + v₀t + at²/2

x = v/3.3²/2

x = 3v/2

II - Uniform

s = s₀ + vt

s = 200

s₀ = x

v = v

t = 19.30 - 3 = 16.30s

s = s₀ + vt

200 = x + v*16.3

100 = x + 16.3v

As x = 3v/2

200 = 3v/2 + 16.3v

200 = 1.5v + 16.3v

200 = 17.8v

v = 11.24m/s

a = v/3 = 3.75 m/s²

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Answer: NP = 9 inches

Step-by-step explanation:

In a parallelogram, the opposite sides are equal.

If the length of the longer side is MN, it means that the length of the two opposite longer sides is 2MN.

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If the perimeter of MNPQ is 68 inches, it means that

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Step-by-step explanation:

Given:

The figure constructed shows a rectangular prism made up of small wooden cubes of length \frac{1}{8}\ ft.

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Solution:

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