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anonymous

  • one year ago

A runner on a 100-m dash starts from rest and then accelerates to his maximum speed of 10.0 m/s to 4.00 s. He then continues with his speed until the end of the race. a. How many seconds after he starts does the runner finish the race? b. How far from the starting line is the runner at t = 4.00 s c. What is the average speed of the runner for the entire race d. If, upon the crossing the finish line, the runner immediately decelerates at a rate of 2.00 m/s^2, how far from the finish line will the runner stop?

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  1. anonymous
    • one year ago
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    I am the physicist here

  2. anonymous
    • one year ago
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    Do you want the full solutions or step by step walk you through?

  3. anonymous
    • one year ago
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    Assuming his acceleration is constant: \[a = \frac{ \left(10\frac{ m }{s}\right)-\left(0\frac{ m }{s}\right)}{4s - 0s} = 2.5 \frac{ m }{ s ^{2} }\] Since his initial position and velocity was 0 m and 0 m/s respectively; by integrating the equation twice to obtain: \[x(t) = \frac{ 1 }{ 2 } \left( 2.5 \frac{ m }{ s ^{2} }\right) t^{2}\] Using this equation, we find the distance he traveled while accelerating by: \[x(4s) = \frac{ 1 }{ 2 } \left( 2.5 \frac{ m }{ s ^{2} }\right) (4s)^{2} = 20m\] He then proceeds to run at 10 m/s for the remaining 80m left in the sprint, which takes a further: \[t = \frac{ 80m }{ 10 m s^{-1} } = 8 s\] Therefore the total time for the dash is 12s.

  4. anonymous
    • one year ago
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    You seem quite sound there my friend good job

  5. anonymous
    • one year ago
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    Are you taking physics at uni? I am 2nd year physics

  6. anonymous
    • one year ago
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    Yea I actually recently finished my BS and have started a graduate program. I just found this site and am getting some practice at tutoring the basics basics since I am trying to get a job with the Princeton review to help float the grad school costs.

  7. anonymous
    • one year ago
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    Awesome! So physics BS from Priceton university?

  8. anonymous
    • one year ago
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    Right now I am double majoring in medicine and physics and I am pursuing to be a medical physicist however sophistication on either subject requires sacrifice of the other to some extent.

  9. anonymous
    • one year ago
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    No unfortunately not that prestigious, FSU.

  10. anonymous
    • one year ago
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    I assume you were pure physics major then?

  11. anonymous
    • one year ago
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    Physics and mathematics

  12. anonymous
    • one year ago
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    That's a mutualism on both.

  13. anonymous
    • one year ago
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    Medicine sacrifices physics and whereas physics promotes it... in terms of logical reasoning

  14. anonymous
    • one year ago
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    Can't stand the universal hatred for physics amongst people

  15. anonymous
    • one year ago
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    Totally... hang on for a minute I am typing up the remainder for this guy :)

  16. anonymous
    • one year ago
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    Hey I think sig figs is set to 3 so 12.0 s would be more appropriate?

  17. anonymous
    • one year ago
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    I think it's intentional that set it that way... Otherwise I would take 100m as having 1 sig fig which would round the answer to 10.... but deliberate use of .00 implies against it

  18. anonymous
    • one year ago
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    my prof used to always pick me on sig figs until they were stuck as nightmares every now and then

  19. anonymous
    • one year ago
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    The average speed of the runner throughout the race is calculated as follows: Given the time taken for completion of the race;12.0 seconds Total distance covered;100m 100m/12.0s=8.33m/s The average speed of the runner throughout the race is 8.33m/s. *Note that this is a scalar quantity for the direction is unspecified.

  20. anonymous
    • one year ago
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    Given the speed at which the runner crosses the line;10m/s The rate of negative acceleration(be careful with the use of deceleration);-2.00m/s^2 The time taken for the runner to stop is calculated as follows 10m/s-2.00m/s^2(t)=0 -2.00m/s^2(t)=-10m/s t=5.00s Therefore the time taken for the runner to come to the state of rest is 5.00s given the sig figs of 3

  21. anonymous
    • one year ago
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    To calculate the distance from the point at which negative acceleration was applied, we consider the initial velocity as well as the negative acceleration by which the initial velocity was reduced to zero. Therefore, 10m/s(5s)-0.5(2.00m/s^2)(5s)^2=50m-25m=25.0m Therefore the distance covered following negative acceleration upon crossing the line is 25.0m

  22. anonymous
    • one year ago
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    The last question is two fold...

  23. anonymous
    • one year ago
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    The average speed of the sprinter can be found by dividing total distance by total time: \[v_{avg} = \frac{ 100 m }{ 12s } = 8.33 \frac{ m }{s}\] The last part isn't to bad. Take the initial velocity to be: \[v_{0} = 10 \frac{ m }{ s }\] And consider the finish line to be the new 0 point (aka located at 0 m). Since the runners acceleration is again constant, but has magnitude (negative because his velocity is decreasing): \[a = -2.0\frac{ m }{ s^{2} }\] Integrating this equation once gives: \[v(t) = -a t + v_{0} \rightarrow t = \frac{v(t) - v_{0}}{a}\] Since his final veloctiy is zero after this interval -> v(T) = 0 \[T = \frac{- v_{0}}{a} = 5s\] Integrating the velocity equation one more time: \[x(t) = \frac{ -1 }{ 2 } a t^{2} + v_{0}t\] And plugging in 5s yields: \[x(5s) = \frac{ 1 }{ 2 } a (5s)^{2} + v_{0}(5s) = -25m + 50m = 25m \]

  24. anonymous
    • one year ago
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    Whew editing is tough in this thing it is quite annoying and time consuming :(

  25. anonymous
    • one year ago
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    Oh you beat me to the punch :p good job

  26. anonymous
    • one year ago
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    Math majors are impressive with equations... wow I am so impressed.

  27. anonymous
    • one year ago
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    I reason with concepts but math majors like every bit of equations.. wow

  28. anonymous
    • one year ago
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    Well dont be too impressed answers are what matter and we both got the same :D

  29. anonymous
    • one year ago
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    I think sig figs be 3

  30. anonymous
    • one year ago
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    otherwise 1... which would render most of our answers totally inaccurate(100m) part

  31. anonymous
    • one year ago
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    Youre probably right I have been quite sloppy with those.... math on numbers seems almost foreign to me now jk

  32. anonymous
    • one year ago
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    but like I said consistency with 3 implies against it

  33. anonymous
    • one year ago
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    Yea I spent all my time trying to get those to look good and be formatted correctly didnt even think of sig figs tbh

  34. anonymous
    • one year ago
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    Yeah your answer is pretty professional. It's a mathematician's physics.

  35. anonymous
    • one year ago
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    Mine is classical old fart physicist's physics

  36. anonymous
    • one year ago
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    JK

  37. anonymous
    • one year ago
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    :D

  38. anonymous
    • one year ago
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    Oh god. Thankyou!! Guyssss sorry took so long to reply. I'm at school lolol thanks again I'll read everything first

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