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Coasting due west on your bicycle at 8.4 m/s, you encounter a sandy patch of road 7.2 m across. When you leave the sandy patch, your speed has been reduced to 6.4 m/s. What is the bicycles acceleration in the sandy patch? assume that the acceleration is constant and that the direction of travel is the positive direction.

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MathPhys

Answer:

-2.1 m/s²

Explanation:

Given:

v₀ = 8.4 m/s

v = 6.4 m/s

Δx = 7.2 m

Find: a

v² = v₀² + 2aΔx

(6.4 m/s)² = (8.4 m/s²) + 2a (7.2 m)

a = -2.1 m/s²

Lanuel

The bicycle's acceleration in the sandy patch is -2.06 [tex]m/s^2[/tex].

Given the following data:

  • Initial velocity = 8.4 m/s.
  • Final velocity = 6.4 m/s.
  • Distance = 7.2 meters.

To find the bicycle's acceleration in the sandy patch, we would use the third equation of motion;

[tex]V^2 = U^2 + 2aS[/tex]

Where:

  • V is the final velocity.
  • U is the initial velocity.
  • a is the acceleration.
  • S is the distance covered.

Substituting the given parameters into the formula, we have;

[tex]6.4^2 = 8.4^2 + 2a(7.2)\\\\40.96 = 70.56 + 14.4a\\\\14.4a = 40.96 - 70.56\\\\14.4a = -29.6\\\\a = \frac{-29.6}{14.4}[/tex]

Acceleration, a = -2.06 [tex]m/s^2[/tex]

Therefore, the bicycle's acceleration in the sandy patch is -2.06 [tex]m/s^2[/tex].

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