Given:
The position and the time graph of a dog is shown in the figure given in the question.
To find:
the position of the dog at t=3 s, velocity of the dog for interval A, and the average velocity of the dog over intervals A and B.
Explanation:
The position of the dog at t=3 s, is,
[tex]y=8\text{ m \lparen from graph\rparen}[/tex]
The final position after interval A is,
[tex]8\text{ m}[/tex]
The initial position before the interval A is,
[tex]5\text{ m}[/tex]
The velocity for the interval A is,
[tex]\begin{gathered} \frac{The\text{ displacement}}{The\text{ time interval}} \\ =\frac{8-5}{2-0} \\ =\frac{3}{2} \\ =1.5\text{ m/s} \end{gathered}[/tex]
The change in position at interval B is 0, so the velocity at interval B is 0.
The average velocity over the interval A and B is,
[tex]\begin{gathered} \frac{The\text{ velocity at interval A+The velocity at interval B}}{2} \\ =\frac{(1.5)+\text{0}}{2} \\ =\frac{1.5}{2} \\ =0.75\text{ m/s} \end{gathered}[/tex]
Hence, the position of the particle after the required time is 8 m, The velocity after interval A is 1.5 m/s, and the average velocity over the interval A and B is 0.75 m/s.
