- #36
Medgirl314
- 561
- 2
Is that correct so far?
Medgirl314 said:Is that correct so far?
I think we've gone a bit off track here. First, let me correct the units above, which I didn't think to do before. Both of those numbers are in units of m2/s2. On one side that's from multiplying two speeds, on the other from multiplying an acceleration by a distance. This is not a route to calculating time.Medgirl314 said:70.8 m/s^2=70.4 m.s^2
I realize that's a bit off, but I *think* it's just because of my rounding earlier. Is the answer simply approximately 70.6 m/s? Thanks again for all your help!
and my post #32:someone else suggested we calculate the average
I did not mean to suggest this was a useful way to calculate the time. I was just explaining a connection between the equations.The reference to average velocity was distance = vavg * time. We can rewrite [v2=v02+2aΔx] using an average: vavg * Δv = aΔx.
You found the initial velocity of the rock to be 11.9m/s and so you then said that vavg= 11.9/2 = 5.95m/s. But the value of 11.9m/s has been rounded, so to obtain the correct average velocity you should work with the exact result.Medgirl314 said:Would you mind explaining when I incorrectly rounded vavg? My physics teacher says he's not too worried about how I round.
Yes, and the 'entire decimal answer' is exactly √140.8.Medgirl314 said:Okay, I think I understand what you're saying. You're saying that after I calculated the inital velocity, I should have kept the entire decimal answer, and then worked through our equations without rounding, but then round at the very end, correct?
Is this for a online test? It might say 'Find the initial velocity to X amount of significant figures', in which case you should do so. But for subsequent parts of the same question, always use the exact value.So I should go back to calculating the intial velocity, record the unrounded answer, and then go back and re-do the formula in #8.
Sorry, √140.8 is a surd since you cannot simplify it further. It is not an abbreviation, just a common term. +ve means positive.The only thing I'm not clear on is the abbrevations. What is surd and +ve? Thanks again!
CAF123's point is that if you evaluate some variable as a number, then use that number in further calculations, then any rounding error gets carried into the second calculation. Sometimes this can result in a much bigger error in the final answer than you might expect. It's almost always best to keep the algebraic form as long as you can along the path to each result. Sometimes you get some cancellation.Medgirl314 said:It's not an online test, it's just a homework set, but my physics teacher usually assumes we'll round instead of leaving answers as a square root, as far as I know.This problem could be different. He hasn't explained surds yet, or gotten into leaving answers as square roots, so that may be where some of the confusion is coming from. Would you mind clarifying what equation I need to use at this point?
No, that's much too low. Pls post your working for that.Medgirl314 said:That seems to yield an answer of approximately .27 seconds, which seems far more plausiable than my original answer.
Two problems there. The average speed is only half the launch speed, so 5.9329587897.Medgirl314 said:I used this formula: vavg=distance/time,
Plugged in my numbers:11.86591758=44/t
And got about .27.
Not quite. What time do you think you have calculated here? What time does it ask for?The square root of 55 is approximately 7.42. Is that the answer?
Medgirl314 said:It asks for the total time, and that time is just how long it would take to go up. So the answer would be more like 14.84 seconds.