Example

Question:

The position of an object moving along the x-axis is given by \( x = a + bt^2 \) where \( a = 8.5\,\mathrm{m} \), \( b = 2.5\,\mathrm{m/s^2} \), and \( t \) is measured in seconds.
What is its velocity at \( t = 0\,\mathrm{s} \) and \( t = 2.0\,\mathrm{s} \)?
What is the average velocity between \( t = 2.0\,\mathrm{s} \) and \( t = 4.0\,\mathrm{s} \)?

Solution:

In notation of differential calculus, the velocity is \[ v = \frac{dx}{dt} = \frac{d}{dt}(a + bt^2) = 2bt \] At \( t = 0\,\mathrm{s} \): \[ v = 2b \times 0 = 0~\mathrm{m/s} \] At \( t = 2.0\,\mathrm{s} \): \[ v = 2b \times 2.0 = 5.0~\mathrm{m/s} \]
The average velocity between \( t = 2.0\,\mathrm{s} \) and \( t = 4.0\,\mathrm{s} \) is \[ \text{Average velocity} = \frac{x(4.0) - x(2.0)}{4.0 - 2.0} \] Calculate positions: \[ x(4.0) = a + b \times 16 = 8.5 + 2.5 \times 16 = 8.5 + 40 = 48.5~\mathrm{m} \] \[ x(2.0) = a + b \times 4 = 8.5 + 2.5 \times 4 = 8.5 + 10 = 18.5~\mathrm{m} \] So, \[ \text{Average velocity} = \frac{48.5 - 18.5}{2} = \frac{30}{2} = 15~\mathrm{m/s} \]