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BloomLocke367

  • one year ago

Help? I'm pretty sure I have it, but not 100%

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  1. BloomLocke367
    • one year ago
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    The highest speed achieved by a standard nonracing sports car is \(3.50\times 10^2~km/hr\). Assuming that the car accelerates at 4.00m/s, how long would it take this car to reach its maximum speed if it is initially at rest? What distance would the car travel during this time?

  2. BloomLocke367
    • one year ago
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    I for sure have the first question (and work to prove it) but I'm not sure if I need to do one more step on the last question. The units are throwing me off.

  3. BloomLocke367
    • one year ago
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    @Nnesha, can you help?

  4. BloomLocke367
    • one year ago
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    @mathmate

  5. BloomLocke367
    • one year ago
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    Can you help?

  6. mathmate
    • one year ago
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    Yes, units are the key!

  7. mathmate
    • one year ago
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    I prefer to work with metres/second, because these are standard / basic units.

  8. mathmate
    • one year ago
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    so 3.5*10^2 km/hr = 350 km/hr = 350*1000 m / 3600 s. = 875/9 m/s (exact conversion)

  9. BloomLocke367
    • one year ago
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    The first thing I did was convert km/hr to m/s by doing \(\Huge(\frac{350km}{1hr})(\frac{1hr}{60min})(\frac{1min}{60sec})(\frac{1000m}{1km})\) and I got 97.2m/s

  10. BloomLocke367
    • one year ago
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    that's with keeping the right amount of significant figures.

  11. mathmate
    • one year ago
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    Yes, 97.2 m/s is correct to one place after decimal. I prefer to keep the exact values until the end, if possible.

  12. mathmate
    • one year ago
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    So what did you get for the time to accelerate?

  13. BloomLocke367
    • one year ago
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    okay then I used the formula \(v=v_0+at\) and I got 24.3s for the time.

  14. mathmate
    • one year ago
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    Yep, that's what I got too, 24.306 (if you want to be more precise). Which kinematics equation would you use to find the distance?

  15. BloomLocke367
    • one year ago
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    I used \(v^2=v_0^2+2a(x-x_0)\)

  16. mathmate
    • one year ago
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    That's good! So what do you get with the distance (x-x0)?

  17. mathmate
    • one year ago
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    The term v^2 will exaggerate the round-off errors, so be sure to keep more significant figures.

  18. BloomLocke367
    • one year ago
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    welllllll, that's my problem. I got all the way down to my last step where I have to divide \(9447.84m^2/s^2\) and \(8.00m/s^2\). I got lost with the units at this point. I'm not sure what I'm left with after I divided

  19. mathmate
    • one year ago
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    Units can be manipulated exactly as variables, as you can see below. In physics (and chemistry), this is a very important technique to keep track of the units (dimensions) of the answer. With some practice, you'll be very good at it. 9447.84m2/s2 / 8.00m/s2 =9447.84 / 8 * (m^2/m) / (s^2/s^2) =1180.98 m ( I get 1181.52 m if I keep sufficient significant figures). (gtg, will be back later)

  20. BloomLocke367
    • one year ago
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    I got the same number and I thought it was only m left but I wanted to be sure I didn't need to take the square root. Thanks for the help :)

  21. BloomLocke367
    • one year ago
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    wait wait wait. How did you get 1181.52m?

  22. BloomLocke367
    • one year ago
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    Shouldn't there be only 3 significant fig?

  23. mathmate
    • one year ago
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    Yes, the answer should give only 3 significant figures. To make sure the third figure is correct, you would want to carry out your calculations keeping at least 4 or even 5 if there are powers involved. Then round to three for the answer.

  24. mathmate
    • one year ago
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    By solving (875/9)^2=0^2+2(4.00)S S=1181.52...

  25. BloomLocke367
    • one year ago
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    I got 1180.98... I'm so confused.

  26. mathmate
    • one year ago
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    94.2^2=9447.84 (rounded too early) (875/9)^2=9452.16.... (exact)

  27. mathmate
    • one year ago
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    As I said earlier, whenever a number is raised to a power (like squared), the round-off error is magnified. Squaring will double the round off error.

  28. BloomLocke367
    • one year ago
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    ohhhhh okay.

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