A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

YumYum247

  • one year ago

how do i plot a velocity time graph from an average velocity of an object?

  • This Question is Closed
  1. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    if a runner accelerated along the sidewalk at 0.5m/sec2 for 20 sec. Assume that the runner started from rest position.....

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You will plot velocity on the vertical (y) axis and the time on the horizontal (x) axis. All you have to do is plot the values and connect all of them by a line. Make sure velocity is given in meters per second (m/s) and time in seconds (s) unless instructed otherwise.

  3. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    can you please show me how do i make the data table so i can plot the points neatly...

  4. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1438454610769:dw|

  5. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    how do i use the average velocity given in the question to plot the points on the graph.

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Velocity (in m/s) is [distance in metres] divided by [time in seconds.]

  7. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so v = d/t

  8. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    but how's that gunna help me?

  9. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    if i want to know how far did the runner get in 7 sec, how do i figure that out???

  10. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @oldrin.bataku

  11. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if the acceleration is constant \(a\) then the total change in velocity from the start until time \(t\) is just \(v-v_0=at\); in our case, we starta t rest so \(v_0=0\)

  12. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes the accelaeration is constant.

  13. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so the velocity is given by \(v=at\), so in terms of a velocity vs time plot we see that the graph is a simple line with slope \(a\) passing through the origin

  14. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i just need to plot how much distance he covered from rest to 20sec.

  15. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well, if we have a velocity vs time plot, then the distance covered between \(t=t_1\) and \(t=t_2\) is just the total area bounded by our velocity graph between those points:

  16. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1438456604986:dw|

  17. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    can you please just do one for me, i'll do the rest myself.... can you figure out how much distance the runner covered in let's say 7 seconds????

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1438456642313:dw|

  19. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so the distance they go between \(t_1,t_2\) is just the area under the graph between \(t_1,t_2\).

  20. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    the area under the triangle is the displacement of the object....i get that...but

  21. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    did you use Tf - Ti = a X t ?

  22. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    because i need to make a data table to record all the points until 20 second and plot them on the velocity time graph.

  23. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay, so here we have \(t_1=0\), \(t_2=7\), and our equation is again \(v(t)=0.5t\). so our total displacement is: $$A=\frac12 bh$$here our base is \(t_2=7\) since \(t_1=0\) and our height is \(v(t_2)=v(7)=3.5\) so: $$A=\frac12(7)(3.5)=12.25$$

  24. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you could also have done that work using an integral instead of manually finding teh area of a triangle

  25. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    it doesn't look right because i have the answer key and the maximum displacement the runner makes under 20 sec is 10 meters.... i'm willing to show you the answer tho but i don't know how they got those points.....in the first place.

  26. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i dont know what points youre talking about

  27. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1438457170341:dw|

  28. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    this is how it looks......

  29. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the kinematic equation gives \(x=x_0+v_0 t+\frac12 at^2\), here \(a=0.5,t=7\) and \(x_0=v_0=0\) so \(x=\frac12\cdot0.5\cdot7^2=12.25\)

  30. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    actually, no it doesn't; you've misread it. that's the *velocity* against time, so it's going 10 meters *per second* in velocity after 20 seconds of time

  31. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i'm sorry it's a velocity time graph, the velocity/speed of the runner goes up to 10m/s under 20sec

  32. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes, and that doesn't contradict the fact the runner goes 12.25 meters in 7 seconds.

  33. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ok

  34. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    his speed after 7 seconds is only 3.5 m/s as I stated before

  35. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    with an average speed of \((3.5-0)/2=1.75\) m/s, which is indeed correct: $$\frac{12.25}7=1.75$$ m/s

  36. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ok what formula did you use plase?????? :(

  37. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ok so i need to make a data table of this situation as drawn above, so what equation do i use to plot the speed of the runner every second????

  38. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    since the velocity vs time graph is a line, the average velocity between \(t_1=0,t_2=7\) s is just the midpoint, hence \((3.5-0)/2=1.75\) m/s. this matches the alternative calculation of the average velocity as the total displacement \(12.25\) m divided by the time elapsed \(t_2-t_1=7 \) s, since \(12.25/7=1.75\) m/s as well

  39. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the equation for the velocity at time \(t\) is just \(v=0.5t\), since the acceleration is constant. this is the straight line we've been graphing

  40. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ok let me try on my own and i'll ask you to check my work. Thank you!!! :")

  41. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    omg i see how you got 10m....LOL

  42. YumYum247
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ok one more thing, if the average velocity of an object is 0.5m/sec2 how quick or how fast will it accelerate every second.

  43. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.