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Explain Newton's First Law of Motion and calculate the acceleration and velocity of an object given specific parameters.

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2 parts
b(i)

0.0174 m/s2\text{0.0174 m/s}^2

b(ii)

0.122 m/s\text{0.122 m/s}

Step-by-step solution

5 steps

a) The law of inertia states that an object at rest will stay at rest, and an object in motion will stay in motion with the same speed and in the same direction, unless acted upon by an unbalanced external force. This is also known as Newton's First Law of Motion.

Part b(i)· 3 steps

Determine its acceleration.

Given:
Initial velocity, u=0 m/su = 0 \text{ m/s} (starts from rest)
Distance, s=49 ms = 49 \text{ m}
Time, t=75 st = 75 \text{ s}

  1. Use the kinematic equation relating distance, initial velocity, time, and acceleration.

    s=ut+12at2s = ut + \frac{1}{2}at^2

  2. Substitute the given values into the equation.

    49 m=(0 m/s)(75 s)+12a(75 s)249 \text{ m} = (0 \text{ m/s})(75 \text{ s}) + \frac{1}{2}a(75 \text{ s})^2 49 m=0+12a(5625 s2)49 \text{ m} = 0 + \frac{1}{2}a(5625 \text{ s}^2) 49 m=2812.5 s2⋅a49 \text{ m} = 2812.5 \text{ s}^2 \cdot a

  3. Solve for acceleration aa.

    a=49 m2812.5 s2a = \frac{49 \text{ m}}{2812.5 \text{ s}^2} a≈0.01742 m/s2a \approx 0.01742 \text{ m/s}^2

    The acceleration is 0.0174 m/s2\boxed{\text{0.0174 m/s}^2}.

Part b(ii)

0.122 m/s\text{0.122 m/s}

Determine its velocity after 7 seconds.

  1. Use the kinematic equation relating final velocity, initial velocity, acceleration, and time.

    v=u+atv = u + at

  2. Substitute the initial velocity, calculated acceleration, and new time (t=7 st = 7 \text{ s}) into the equation.

    v=0 m/s+(0.01742 m/s2)(7 s)v = 0 \text{ m/s} + (0.01742 \text{ m/s}^2)(7 \text{ s}) v=0.12194 m/sv = 0.12194 \text{ m/s}

    The velocity after 7 seconds is 0.122 m/s\boxed{\text{0.122 m/s}}.

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