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A body moving with uniform acceleration has a velocity of –11 cm/s when its x coordinate is 3.00 cm. If its x coordinate 2 s later is – 5 cm, what is the magnitude in cm/s2 of its acceleration? KO0006?
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A body moving with uniform acceleration has a velocity of –11 cm/s whe...
Given:

  • Initial velocity (u) = -11 cm/s

  • Initial position (x0) = 3.00 cm

  • Final position (x) = -5 cm

  • Time taken (t) = 2 s



Formula:

  • Acceleration (a) = (v - u)/t

  • Final velocity (v)² = u² + 2a(x - x0)



Solution:

  • Using the first formula, acceleration can be calculated as:


    • a = (v - u)/t

    • a = (-5 - (-11))/2

    • a = 3 cm/s²



  • Using the second formula, final velocity can be calculated as:


    • v² = u² + 2a(x - x0)

    • v² = (-11)² + 2(3)(-5 - 3)

    • v² = 121 - 48

    • v = -7 cm/s



  • Therefore, the magnitude of acceleration is 3 cm/s².



Explanation:

  • The problem involves a body moving with uniform acceleration, which means that the acceleration is constant over time.

  • The initial velocity and position are given, and the final position is also known after a certain amount of time.

  • To find the acceleration, we use the formula a = (v - u)/t, where v is the final velocity, u is the initial velocity, and t is the time taken.

  • To find the final velocity, we use the formula v² = u² + 2a(x - x0), where x is the final position, x0 is the initial position, and a is the acceleration.

  • By solving the two equations for acceleration and final velocity, we can find the required values.

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A body moving with uniform acceleration has a velocity of –11 cm/s when its x coordinate is 3.00 cm. If its x coordinate 2 s later is – 5 cm, what is the magnitude in cm/s2 of its acceleration? KO0006?
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