A person run 100 m in 6 s and then move back for 70 m in 4 s.

Determine the average speed and the average velocity of the person.

A person run 100 m in 6 s and then move back for 70 m in 4 s.

Determine the average speed and the average velocity of the person.

Solution: 

Distance traveled is \(170 \text{ m}\) and the displacement is \(30 \text{ m}\).

\(\text{Average speed} = \dfrac{170}{10} = \bbox[5px, border: 2px solid magenta] {17 \text{ m/s}}\)

\(\text{Average velocity} = \dfrac{30}{10} = \bbox[5px, border: 2px solid magenta] {3 \text{ m/s}}\)

A particle moves from A to B in 6 seconds and continue its moving to C in 4 seconds, as graph below (unit in meter).

Determine:

A. Average speed of the particle from A to C

B. Average velocity of the particle from A to C

A particle moves from A to B in 6 seconds and continue its moving to C in 4 seconds, as graph below (unit in meter).

Determine:

A. Average speed of the particle from A to C

B. Average velocity of the particle from A to C

Solution: 

A. Average speed of the particle from A to C

Distance covered by particle is AB + BC = \(4 + 3 = 7 \text{ m}\)

Total time from A to C = \(6 + 4 = 10 \text{ s}\)

\(\text{Average speed} = \dfrac{\text{distance}}{\text{time}} = \dfrac{7}{10} = \bbox[5px, border: 2px solid magenta] {0,7 \text{ m/s}}\)

B. Average velocity of the particle from A to C

Initial and final position of the particle are A and C. Then displacement of the particle is AC. By Phytagoras theorem:

\begin{equation*} \begin{split} \text{AC}^2 & = \text{AB}^2 + \text{BC}^2 \\\\ \text{AC}^2 & = 3^2 + 4^2 \\\\ \text{AC}^2 & = 25 \\\\ \text{AC} & = \pm \sqrt{25} \\\\ \text{AC} & = \pm 5 \end{split} \end{equation*}

Since length value must be positive, take \(\text{AC} = +5\)

\(\text{Average velocity} = \dfrac{\text{displacement}}{\text{time}} = \bbox[5px, border: 2px solid magenta] {\dfrac{5}{10} = 0,5 \text{ m/s}}\)

A particle moves 1 round of circle with radius 10 cm in 5 seconds. Determine:

A. Average speed

B. Average velocity

A particle moves 1 round of circle with radius 10 cm in 5 seconds. Determine:

A. Average speed

B. Average velocity

Solution: 

Distance covered by the particle is perimeter of the circle.

\(2 \:.\: \pi \:.\: r = 2 \:.\: 3,14 \:.\: 10 =  62.8 \text{ m}\)

\(\text{Average speed} = \dfrac{\text{distance}}{\text{time}} = \dfrac{62,8}{5} = \bbox[5px, border: 2px solid magenta] {12,56 \text{ m/s}}\)

Displacement of the particle is \(0\) because the particle is back to its initial position.

\(\text{Average velocity} = \dfrac{\text{displacement}}{\text{time}} = \bbox[5px, border: 2px solid magenta] {0}\)

A particle moves in one direction follows:

\(x = 2t^2 + 5t – 2\)

where \(x\) is the position of the particle in meter and \(t\) is time traveled in seconds.

Determine its average velocity in the first 10 seconds of its movement.

A particle moves in one direction follows:

\(x = 2t^2 + 5t – 2\)

where \(x\) is the position of the particle in meter and \(t\) is time traveled in seconds.

Determine its average velocity in the first 10 seconds of its movement.

Solution: 

Displacement

\begin{equation*} \begin{split} \Delta x & = x_{(10)} – x_{(0)} \\\\ \Delta x & = [2(10)^2 + 5(10) – 2] – [2(0)^2 + 5(0) – 2] \\\\ \Delta x & = [200 + 50 – 2] – [-2] \\\\ \Delta x & = \bbox[5px, border: 2px solid magenta] {250 \text{ m}} \end{split} \end{equation*}

Average velocity

\begin{equation*} \begin{split} \bar{v} & = \frac{\text{displacement}}{\text{time}} \\\\ \bar{v} & = \frac{250}{10} \\\\ \bar{v} & = \bbox[5px, border: 2px solid magenta] {25 \text{ m/s}} \end{split} \end{equation*}