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An object is moving at -6.90 m/s. It accelerates uniformly over a distance of 3.20 m. Its final speed is +9.70 m/s. What is its acceleration?

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Final answer:

The acceleration of an object that accelerates uniformly from -6.90 m/s to +9.70 m/s over a distance of 3.20 m is calculated using the kinematic equation and results in an acceleration of 7.28 m/s^2.

Step-by-step explanation:

To calculate the acceleration of an object given its initial velocity, final velocity, and the distance over which it accelerates, we can use the kinematic equation:

v^2 = u^2 + 2as

Where:

  • v is the final velocity,
  • u is the initial velocity,
  • a is the acceleration,
  • s is the distance over which the acceleration occurred.

Given that the initial velocity u = -6.90 m/s, the final velocity v = +9.70 m/s, and the distance s = 3.20 m, we can rearrange the formula to solve for a:

a = (v^2 - u^2) / (2s)

Substituting the provided values into the rearranged formula:

a = (9.70^2 - (-6.90)^2) / (2 * 3.20)

After performing the calculations:

a = (94.09 - 47.61) / 6.40 = 7.28 m/s^2

The acceleration of the object is 7.28 m/s^2.

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