Question

The position of a particle moving along the \(x\) -axis is given by \(x=\left(11+14 t-2.0 t^{2}\right),\) where \(t\) is in seconds and \(x\) is in meters. What is the average velocity during the time interval from \(t=1.0 \mathrm{~s}\) to \(t=4.0 \mathrm{~s} ?\)

Short answer

Answer: The average velocity during the time interval from \(t=1.0\text{ s}\) to \(t=4.0\text{ s}\) is \(4.0 \frac{\text{m}}{\text{s}}\).

Step-by-step solution

Calculate Position at the Starting and Ending of the Time Interval

Using the position function, find the position of the particle at \(t=1.0s\) and \(t=4.0s\): \(x(1.0) = 11 + 14(1.0) - 2.0(1.0)^2\) \(x(4.0) = 11 + 14(4.0) - 2.0(4.0)^2\)

Compute Positions

Calculate the positions at the given times: \(x(1.0) = 11 + 14 - 2 = 23\text{ m}\) \(x(4.0) = 11 + 56 - 32 = 35\text{ m}\)

Find Displacement

Subtract the initial position from the final position to find the displacement: \(\Delta x = x(4.0) - x(1.0) = 35 - 23 = 12\text{ m}\)

Calculate Time Interval

Find the time interval between the given times: \(\Delta t = 4 - 1 = 3\text{ s}\)

Compute Average Velocity

Divide the displacement by the time interval to find the average velocity: \(v_\text{avg} = \frac{\Delta x}{\Delta t} = \frac{12}{3} = 4.0 \frac{\text{m}}{\text{s}}\) The average velocity during the time interval from \(t=1.0\text{ s}\) to \(t=4.0\text{ s}\) is \(4.0 \frac{\text{m}}{\text{s}}\).