Written Problem: Plotting Kinematics

Harley the ferret is racing through a long straight pipe running from right to left as part of the national ferreting championships. Harley's run goes like this:

-He travels at a constant rate of 0.80 m/s to the left for the first 2.5 s

-Then he travels at 0.10 m/s to the right for 5.0 s

-He stops for another 2.5 s to have a scratch before finally travelling to the left, towards the finish line, at a constant 0.25 m/s for 10 s. He crosses the finish line at the end of this time.

a) Draw a velocity versus time plot for the ferret.

b) Draw a position versus time plot for the ferret.

c) Draw a distance versus time plot for the ferret.

d) How long is the pipe?

e) What is the average velocity of the ferret?

f) What is the average speed of the ferret?

g) Draw an acceleration versus time plot for the ferret. Say something about what happens between time intervals (for example: what happens between the interval 0-2.5 seconds and the interval 2.5-5 seconds?)

Respuesta :

Answer:

d) [tex]l=5\ m[/tex]

e) [tex]v_{avg}=0.25\ m.s^{-1}[/tex]

f) [tex]s_{avg}=v_{avg}=0.25\ m.s^{-1}[/tex]

Explanation:

Given:

  • Duration of the first interval of time, [tex]t_1=2.5\ s[/tex]
  • speed in the first interval towards left, [tex]v_1=0.8\ m.s^{-1}[/tex]
  • duration of the second interval, [tex]t_2=5\ s[/tex]
  • speed of the second interval towards right, [tex]v_2=-0.1\ m.s^{-1}[/tex]
  • duration of stopping, [tex]t_0=2.5\ s[/tex]
  • duration of final interval of the race, [tex]t_3=10\ s[/tex]
  • speed towards left in the final interval of the race, [tex]v_3=0.25\ m.s^{-1}[/tex]

d)

Now, the length of the pipe will be the total displacement of the racer.

Then the length of the pipe:

[tex]l=v_1.t_1+v_2.t_2+v_3.t_3[/tex]

[tex]l=0.8\times 2.5-0.1\times 5+0.25\times 10[/tex]

[tex]l=4\ m[/tex]

e)

Average velocity is the total displacement travelled divided by the total time taken to cover the displacement.

so,

[tex]v_{avg}=\frac{l}{(t_1+t_2+t_0+t_3)}[/tex]

[tex]v_{avg}=\frac{4}{(2.5+5+2.5+10)}[/tex]

[tex]v_{avg}=0.2\ m.s^{-1}[/tex]

f)

The average speed is equal to the total distance covered by the racer divided by the total time taken by the racer.

[tex]s_{avg}=\frac{v_1.t_1+v_2.t_2+v_3.t_3}{t_1+t_2+t_0+t_3}[/tex]

[tex]s_{avg}=\frac{0.8\times 2.5+0.1\times 5+0.25\times 10}{2.5+5+2.5+10}[/tex]

[tex]s_{avg}=0.25\ m.s^{-1}[/tex]

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