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An object starts from rest and covers a distance of 18 m in 3.0 s. Calculate its acceleration.

Physics

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Answer

4 m/s24 \text{ m/s}^2

Step-by-step solution

3 steps
  1. Identify the given kinematic information.

    The object starts from rest, so its initial velocity is u=0 m/su = 0 \text{ m/s}.
    The distance covered is s=18 ms = 18 \text{ m}.
    The time taken is t=3.0 st = 3.0 \text{ s}.

  2. Choose the appropriate kinematic equation.

    To find the acceleration (aa) given uu, ss, and tt, we use the kinematic equation:

    s=ut+12at2s = ut + \frac{1}{2}at^2
  3. Substitute the known values into the equation and solve for acceleration.

    18 m=(0 m/s)(3.0 s)+12a(3.0 s)218 \text{ m} = (0 \text{ m/s})(3.0 \text{ s}) + \frac{1}{2}a(3.0 \text{ s})^2 18=0+12a(9.0)18 = 0 + \frac{1}{2}a(9.0) 18=4.5a18 = 4.5a a=184.5a = \frac{18}{4.5} a=4 m/s2a = 4 \text{ m/s}^2

    The acceleration of the object is 4 m/s24 \text{ m/s}^2.

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