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calculusxy

  • one year ago

1. How far will an object move in one second if its average speed is 5 m/s? A. 5 meters 2. How far will a freely falling object have fallen from a position of rest when its instantaneous speed is 10 m/s? A. 75 meters 3. An object dropped from rest and falls freely. After 6 seconds, calculate its instantaneous speed, average speed, and distance fallen. A. Instantaneous speed: 58.8 m/s Average Speed : 53.9 m/s Distance : 313.6 meters 4. If a freely falling rock were equipped with an odometer, would the readings for the distance fallen each second stay the same...

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  1. calculusxy
    • one year ago
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    ...increase with tie, or decrease with time? A. The instantaneous speed would increase so the distance would increase as well.

  2. calculusxy
    • one year ago
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    @jim_thompson5910

  3. calculusxy
    • one year ago
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    Can you make sure the answers?

  4. ybarrap
    • one year ago
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    A. Is correct because $$ 5 \frac{m}{\cancel{s}}\times 1~\cancel{s}=5~m $$ Notice how the seconds cancel each other out and you are left with just meters ,\(m\)?

  5. calculusxy
    • one year ago
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    Yeah

  6. ybarrap
    • one year ago
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    For 2 you need to use the following equation $$ v^2 = u^2 + 2as $$ where u is the initial velocity, a is the acceleration of gravity, s is the distance traveled and v is the final velocity: https://en.wikipedia.org/wiki/Equations_of_motion#Kinematic_equations_for_one_particle So $$ 10^2=0^2+2s\times 9.81\\ \implies s=\frac{100}{2\times 9.81}=5~m $$ Does this make sense?

  7. calculusxy
    • one year ago
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    Answer would be 5M?

  8. ybarrap
    • one year ago
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    Yes

  9. calculusxy
    • one year ago
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    how would it be 5m if it is travelling 10m/s

  10. ybarrap
    • one year ago
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    |dw:1436316669347:dw| (ignore the t=1/2 seconds above) It takes $$ v/a=10/9.81=1.01\text{ second} $$ to go from 0 to 10 m/s In that time period, it is accelerating. If it were going a constant speed of 10 m/s then is 1 second it would have traveled 10 m, but it didn't START at 10 m/s, it started at 0 m/s. Right?

  11. calculusxy
    • one year ago
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    yes

  12. ybarrap
    • one year ago
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    For # 3 Instantaneous speed: $$ v=at\\ v=9.8\times 6=58.8~m/s $$ Average speed: $$ \frac{1}{6}\int_0^6 at~dt=\frac{1}{6}a\frac{t^2}{2}|_0^6=\frac{9.8\times 6^2}{12}=29.4~m/s $$ distance $$ s=\frac{1}{2}at^2=\frac{1}{2}9.8(6)^2=176.4~m $$ Does this make sense?

  13. calculusxy
    • one year ago
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    Sorry but I am just a rising eighth grader so i don't understand the average speed part.

  14. ybarrap
    • one year ago
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    How did you determine average?

  15. ybarrap
    • one year ago
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    You can graph it, that's another way, want to try that?

  16. calculusxy
    • one year ago
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    what i did was: 5 sec = 49 m/s 6 sec = 58.8 m/s 49 + 58.8 / 2 = 53.9

  17. ybarrap
    • one year ago
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    Why did you use 5 seconds here?

  18. calculusxy
    • one year ago
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    well that's what my teacher said for me to use

  19. calculusxy
    • one year ago
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    2 seconds => average speed 19.6 + 9.8 /2 = 14.7 m/s

  20. calculusxy
    • one year ago
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    9.8 (1st second) 19.6 (2nd second)

  21. ybarrap
    • one year ago
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    Yes, that is a good approach, keep going

  22. calculusxy
    • one year ago
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    so 53.9 is correct?

  23. ybarrap
    • one year ago
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    It will be an estimate... but if you want a formula, use: $$ \frac{v_{final}-v_{initial}}{2}=\frac{v_{final}-0}{2}=\frac{a t}{2} $$ Because the initial velocity is 0 and final velocity = \(at\).

  24. calculusxy
    • one year ago
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    a = ?

  25. ybarrap
    • one year ago
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    Average speed should be 9.8*6/2. a is the acceleration of gravity = 9.8 m/s^2

  26. calculusxy
    • one year ago
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    29.4

  27. ybarrap
    • one year ago
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    yes

  28. calculusxy
    • one year ago
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    okay

  29. calculusxy
    • one year ago
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    next

  30. ybarrap
    • one year ago
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    So did you get the distance traveled?

  31. calculusxy
    • one year ago
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    for #4?

  32. ybarrap
    • one year ago
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    #3

  33. calculusxy
    • one year ago
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    yeah

  34. ybarrap
    • one year ago
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    176.4 m ?

  35. calculusxy
    • one year ago
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    yes

  36. ybarrap
    • one year ago
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    Ok for the final one...

  37. ybarrap
    • one year ago
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    An odometers measures distance traveled, right?

  38. calculusxy
    • one year ago
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    yes

  39. ybarrap
    • one year ago
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    And acceleration is change in velocity per unit time, so your speed is changing every second, in fact its increasing every second, right?

  40. calculusxy
    • one year ago
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    yes

  41. ybarrap
    • one year ago
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    Ok, so velocity is changing. If in one second you are going 1 m/s and in the next second you are going 2 m/s then the distance you are traveling every second is increasing. So the odometer will show for the 1st second one meter and the odometer will show for the second second, 2 meters. Do you see that?

  42. calculusxy
    • one year ago
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    yes so it's increasing right?

  43. ybarrap
    • one year ago
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    yes, because velocity is increasing. If you were slowing down, then the odometer would do the opposite

  44. calculusxy
    • one year ago
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    thank you so much!

  45. ybarrap
    • one year ago
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    you're welcome!!

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